Question

An advertisement for a popular supermarket chain claims that it has had consistently lower prices than four other full-service supermarkets. As part of a survey conducted by an "independent market basket price-checking company," the average weekly total, based on the prices (in $) of approximately 95 items, is given for two different supermarket chains recorded during 4 consecutive weeks in a particular month. Week Advertiser ($) Competitor ($) 1 254.30 255.94 2 240.71 255.70 3 231.80 255.18 4 234.20 261.24 (a) Is there a significant difference in the average prices for these two different supermarket chains? (Use ? = 0.05. Round your answers to three decimal places.) 1-2. Null and alternative hypotheses. Multiple Choice, choose one of the following: H0: μd = 0 versus Ha: μd < 0 H0: μd = 0 versus Ha: μd > 0 H0: μd ≠ 0 versus Ha: μd = 0 H0: μd < 0 versus Ha: μd > 0 H0: μd = 0 versus Ha: μd ≠ 0 3. Test statistic: t = 4. Rejection region: If the test is one-tailed, enter NONE for the unused region. t > _____ t < _____ 5. Conclusion. Multiple Choice, choose one of the following: H0 is rejected. There is sufficient evidence to indicate that the means are different. H0 is rejected. There is insufficient evidence to indicate that the means are different. H0 is not rejected. There is sufficient evidence to indicate that the means are different. H0 is not rejected. There is insufficient evidence to indicate that the means are different. (b) What is the approximate p-value for the test conducted in part (a)? Multiple Choice, choose one of the following: p-value < 0.010 0.010 < p-value < 0.020 0.020 < p-value < 0.050 0.050 < p-value < 0.100 0.100 < p-value < 0.200 p-value > 0.200 (c) Construct a 99% confidence interval for the difference in the average prices for the two supermarket chains. (Round your answers to two decimal places.) $_______ to $_______ Interpret this interval. Multiple Choice, choose one of the following: Since 0 does not fall in the confidence interval, there is sufficient evidence to indicate that the means are different. Since 0 falls in the confidence interval, there is sufficient evidence to indicate that the means are different. Since 0 does not fall in the confidence interval, there is insufficient evidence to indicate that the means are different. Since 0 falls in the confidence interval, there is insufficient evidence to indicate that the means are different.

          An advertisement for a popular supermarket chain claims that it
has had consistently lower prices than four other full-service
supermarkets. As part of a survey conducted by an "independent
market basket price-checking company," the average weekly total,
based on the prices (in $) of approximately 95 items, is
given for two different supermarket chains recorded during 4
consecutive weeks in a particular month.
Week
Advertiser ($)
Competitor ($)
1
254.30
255.94
2
240.71
255.70
3
231.80
255.18
4
234.20
261.24
(a) Is there a significant difference in the average prices for
these two different supermarket chains? (Use ? =
0.05. Round your answers to three decimal places.)
1-2. Null and alternative hypotheses. Multiple Choice, choose one
of the following:
H0: μd =
0
versus Ha: μd <
0
H0: μd =
0
versus Ha: μd >
0    
H0: μd ≠ 0
versus Ha: μd =
0
H0: μd <
0
versus Ha: μd >
0
H0: μd =
0
versus Ha: μd ≠ 0
3. Test statistic:  t = 
4. Rejection region: If the test is one-tailed, enter NONE for the
unused region.
t > _____
t < _____
5. Conclusion. Multiple Choice, choose one of the
following:
H0 is rejected. There is sufficient
evidence to indicate that the means are different.
H0 is rejected. There is insufficient
evidence to indicate that the means are
different.    
H0 is not rejected. There is sufficient
evidence to indicate that the means are different.
H0 is not rejected. There is
insufficient evidence to indicate that the means are different.
(b) What is the approximate p-value for the test
conducted in part (a)? Multiple Choice, choose one of the
following:
p-value < 0.010
0.010 < p-value <
0.020    
0.020 < p-value < 0.050
0.050 < p-value < 0.100
0.100 < p-value < 0.200
p-value > 0.200
(c) Construct a 99% confidence interval for the difference in the
average prices for the two supermarket chains. (Round your answers
to two decimal places.)
$_______  to $_______ 
Interpret this interval. Multiple Choice, choose one of the
following:
Since 0 does not fall in the confidence interval, there is
sufficient evidence to indicate that the means are different.
Since 0 falls in the confidence interval, there is sufficient
evidence to indicate that the means are
different.    
Since 0 does not fall in the confidence interval, there is
insufficient evidence to indicate that the means are different.
Since 0 falls in the confidence interval, there is insufficient
evidence to indicate that the means are different.
        
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Added by Ramon G.

Elementary Statistics a Step by Step Approach
Elementary Statistics a Step by Step Approach
Allan G. Bluman 9th Edition
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An advertisement for a popular supermarket chain claims that it has had consistently lower prices than four other full-service supermarkets. As part of a survey conducted by an "independent market basket price-checking company," the average weekly total, based on the prices (in $) of approximately 95 items, is given for two different supermarket chains recorded during 4 consecutive weeks in a particular month. Week Advertiser ($) Competitor ($) 1 254.30 255.94 2 240.71 255.70 3 231.80 255.18 4 234.20 261.24 (a) Is there a significant difference in the average prices for these two different supermarket chains? (Use ? = 0.05. Round your answers to three decimal places.) 1-2. Null and alternative hypotheses. Multiple Choice, choose one of the following: H0: μd = 0 versus Ha: μd < 0 H0: μd = 0 versus Ha: μd > 0 H0: μd ≠ 0 versus Ha: μd = 0 H0: μd < 0 versus Ha: μd > 0 H0: μd = 0 versus Ha: μd ≠ 0 3. Test statistic: t = 4. Rejection region: If the test is one-tailed, enter NONE for the unused region. t > _____ t < _____ 5. Conclusion. Multiple Choice, choose one of the following: H0 is rejected. There is sufficient evidence to indicate that the means are different. H0 is rejected. There is insufficient evidence to indicate that the means are different. H0 is not rejected. There is sufficient evidence to indicate that the means are different. H0 is not rejected. There is insufficient evidence to indicate that the means are different. (b) What is the approximate p-value for the test conducted in part (a)? Multiple Choice, choose one of the following: p-value < 0.010 0.010 < p-value < 0.020 0.020 < p-value < 0.050 0.050 < p-value < 0.100 0.100 < p-value < 0.200 p-value > 0.200 (c) Construct a 99% confidence interval for the difference in the average prices for the two supermarket chains. (Round your answers to two decimal places.) $_______ to $_______ Interpret this interval. Multiple Choice, choose one of the following: Since 0 does not fall in the confidence interval, there is sufficient evidence to indicate that the means are different. Since 0 falls in the confidence interval, there is sufficient evidence to indicate that the means are different. Since 0 does not fall in the confidence interval, there is insufficient evidence to indicate that the means are different. Since 0 falls in the confidence interval, there is insufficient evidence to indicate that the means are different.
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Each of the three supermarket chains in the Denver area claims to have the lowest overall prices. As part of an investigative study on supermarket advertising, a local television station conducted a study by randomly selecting nine grocery items. Then, on the same day, an intern was sent to each of the three stores to purchase the nine items. From the receipts, the following data were recorded. At the 0.10 significance level, is there a difference in the mean price for the nine items between the three supermarkets? Item Super's Ralph's Lowblaw's 1 $2.32 $1.25 $1.25 2 $2.40 $1.80 $1.87 3 $2.10 $3.10 $3.10 4 $2.30 $1.87 $1.87 5 $1.21 $1.37 $1.37 6 $4.04 $3.05 $1.72 7 $4.32 $3.52 $2.22 8 $4.15 $3.08 $2.40 9 $5.05 $4.15 $4.21 State the null hypothesis and the alternate hypothesis. For Treatment (Stores): Null hypothesis: H0: μ1 = μ2 = μ3 Alternate hypothesis: H1: There is a difference in the store means. For blocks (Items): Null hypothesis: H0: μ1 = μ2 = ... = μ9 Alternate hypothesis: H1: There is a difference in the item means. What is the decision rule for both? (Round your answers to 2 decimal places.) Complete an ANOVA table. (Round your SS, MS to 3 decimal places, and F to 2 decimal places.) What is your decision regarding the null hypothesis? The decision for the F value (Stores) at 0.10 significance is: Do not reject H0 The decision for the F value (Items) at 0.10 significance is: Do not reject H0 Is there a difference in the item means and in the store means?

David N.

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A market research firm used a sample of individuals to rate the purchase potential of a particular product before and after the individuals saw a new television commercial about the product. The purchase potential ratings were based on a 0 to 10 scale, with higher values indicating a higher purchase potential. The null hypothesis stated that the mean rating "after" would be less than or equal to the mean rating "before." Rejection of this hypothesis would show that the commercial improved the mean purchase potential rating. Use ̑ = 0.05 and the following data to test the hypothesis and comment on the value of the commercial. Individual Purchase Rating After Before 1 6 5 2 6 4 3 7 8 4 4 3 5 3 5 6 9 8 7 7 5 8 6 7 State the null and alternative hypotheses. (Use ̇d = mean rating after − mean rating before.) H0: ̇d = 0 Ha: ̇d ≠ 0 H0: ̇d ≤ 0 Ha: ̇d > 0 H0: ̇d ≠ 0 Ha: ̇d = 0 H0: ̇d ≤ 0 Ha: ̇d = 0

Kari H.

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A 95% confidence interval for μ is calculated as (10, 20). Which of the following would result in a wider confidence interval? When evaluating each option, assume everything stays exactly the same except the one thing mentioned in that option. a. Increasing the sample mean b. Increasing the sample size c. Calculating a 90% confidence interval instead d. Increasing the sample standard deviation The p-value for a hypothesis test for a population mean is 0.01 and the null hypothesis has been rejected. Which one of the following must be true? a. A type I error has occurred. b. A type II error has occurred. c. The significance level of the test is greater than 0.01. d. The significance level of the test is less than 0.01. A trucking company uses the "tyre mania" brand of tires on all of its trucks and wonders if it could save money by changing to a brand of tires that would last longer. They hear of a new brand of tires called "tyre rific" and wish to investigate whether switching brands of tires could save them money. They had 20 trucks outfitted with brand new "tyre mania" tires and 30 trucks fitted with brand new tires from "tyre rific". For each truck, the distance driven was recorded before a tire change was needed due to worn-out tread on one or more of the tires, at which point all tires were changed on the truck. The results were analyzed in R via a 2-sample t-test with a 2-sided alternative hypothesis, and some of the output is summarized below (assume α = 0.05 where necessary). mean sd n tyre rific 70.25634 9.699926 30 tyre mania 65.78654 10.857855 20 Two Sample t-test data: distance by tire t = -1.5219, df = 48, p-value = 0.1346 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: -10.375017 1.435424 Levene's Test for Homogeneity of Variance (center = "median") Df F value Pr(>F) group 1 0.7122 0.4029 48 Let μ1 and μ2 denote the mean kms driven before tire replacement with "tyre mania" and "tyre rific" tires respectively, and σ1² and σ2² be the respective variances. Select the relevant null and alternative hypotheses that have been tested in this analysis. a. H0: μ1 - μ2 = 0 H1: μ1 - μ2 ≠ 0 b. H0: μ1 - μ2 = 0 H1: the tyre rific tires drive further than the tyre mania tires. c. H0: μ2 - μ1 = 0 H1: μ2 - μ1 < 0 d. H0: μ1 - μ2 = 0 H1: μ1 - μ2 > 0

Kari H.


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Transcript

-
00:01 All right, so here we have an advertiser that we're trying to figure out if their groceries are lower than the competitors.
00:10 Okay, so in order to do this, we have some data for four different weeks, okay, and we have a number for the advertiser and the competitor for each week.
00:27 Okay, so what we're going to do is we're actually going to subtract each of this.
00:31 These, okay? and then we're going to test really the differences between the two.
00:38 Okay, so here's all my data, and the last one, 234 .2, and 261 .24.
00:50 And i'm going to subtract each of these in each of these cases, so i really am going to have four data points, which is the differences between each of these.
00:57 So this comes out to negative 1 .64, negative 14 .99, negative 23.
01:04 And negative 27 .04.
01:07 This is the data set that i'm going to do all my analysis on.
01:11 And what i'm really trying to do for the null alternative hypotheses, right, is the null hypothesis i'm trying to see is the mean difference really equal to zero because then they would really be the same, right? versus the alternative would be that the mean difference is less than zero because then that would mean that we really do make less than the competitors.
01:36 Now, it does look like it does just say, is there a significant difference? okay.
01:44 Which is weird because we probably would actually want to check and see if one is greater than the other.
01:54 Okay.
01:55 But this does say that we want to see if there's just a difference.
01:57 So we actually are going to use the not equal to sign for my alternative hypothesis.
02:02 Okay.
02:05 For the next part, we're going to get our test statistic, which is just x bar of the differences minus the mean from the null hypothesis divided by the standard deviation of the differences over the square root of n of the differences, so that's our four, right? so let me put this into my calculator.
02:30 Okay.
02:31 So when i do this, i get a test statistic of negative 2 .3.
02:36 97.
02:39 This is going to be a two -tailed test, meaning that the area here, right, for my alpha is split between the two tails...
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