Question

Consider the following hypotheses: H0: μ = 9,100 HA: μ ≠ 9,100 The population is normally distributed with a population standard deviation of 700. Compute the value of the test statistic and the resulting p-value for each of the following sample results. For each sample, determine if you can "reject/do not reject" the null hypothesis at the 10% significance level. (You may find it useful to reference the appropriate table: z table or t table) (Negative values should be indicated by a minus sign. Round intermediate calculations to at least 4 decimal places. Round "test statistic" values to 2 decimal places and "p-value" to 4 decimal places.) Test statistic p-value a. x̄ = 9,190; n = 105 Reject H0 Do not reject H0 b. x̄ = 9,190; n = 255 Reject H0 Do not reject H0 c. x̄ = 8,830; n = 37 Reject H0 Do not reject H0 d. x̄ = 8,860; n = 37 Do not reject H0 Reject H0

          Consider the following hypotheses:
H0: μ = 9,100
HA: μ ≠ 9,100

The population is normally distributed with a population standard deviation of 700. Compute the value of the test statistic and the resulting p-value for each of the following sample results. For each sample, determine if you can "reject/do not reject" the null hypothesis at the 10% significance level. (You may find it useful to reference the appropriate table: z table or t table) (Negative values should be indicated by a minus sign. Round intermediate calculations to at least 4 decimal places. Round "test statistic" values to 2 decimal places and "p-value" to 4 decimal places.)

Test statistic
p-value

a.
x̄ = 9,190; n = 105
Reject H0 Do not reject H0

b.
x̄ = 9,190; n = 255
Reject H0 Do not reject H0

c.
x̄ = 8,830; n = 37
Reject H0 Do not reject H0

d.
x̄ = 8,860; n = 37
Do not reject H0 Reject H0

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Added by Brittany C.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

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Solved by Expert David Nguyen

07/28/2022

Step 1

Step 1: Understand the Hypotheses and Test Type - Null Hypothesis (H0): μ = 9,100 - Alternative Hypothesis (HA): μ ≠ 9,100 This is a two-tailed test because the alternative hypothesis is not equal to (≠). Show more…

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Consider the following hypotheses: H0: μ = 9,100 HA: μ ≠ 9,100 The population is normally distributed with a population standard deviation of 700. Compute the value of the test statistic and the resulting p-value for each of the following sample results. For each sample, determine if you can "reject/do not reject" the null hypothesis at the 10% significance level. (You may find it useful to reference the appropriate table: z table or t table) (Negative values should be indicated by a minus sign. Round intermediate calculations to at least 4 decimal places. Round "test statistic" values to 2 decimal places and "p-value" to 4 decimal places.) Test statistic p-value a. x̄ = 9,190; n = 105 Reject H0 Do not reject H0 b. x̄ = 9,190; n = 255 Reject H0 Do not reject H0 c. x̄ = 8,830; n = 37 Reject H0 Do not reject H0 d. x̄ = 8,860; n = 37 Do not reject H0 Reject H0

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Transcript

00:01 Hi, i'm david and i'm hitchie happy answering your question.

00:03 Now let me bring up your question here.

00:06 In the question here, we are going to discuss about the hybrid testing.

00:10 Here we are given the null hypothesis when the mean equal to the 9 ,100 versus the only hypothetical hypothesis mean will not equal to the 9 ,100.

00:20 Notice that this could be the 2 -tel test.

00:23 Now we are given the population standard division sigma equal to the 700.

00:31 And then for each question we need to compute the test statistics b value and make the conclusion.

00:39 Now for the question a, we have this will be the x bar and this will be the n.

00:45 So we have the test, it will equal to the z and the formula equal to the x bar.

00:53 We minus the mean over the sigma the one base going to the n.

00:56 And then we have the x bar equal to the 9 -190 minus the n.

01:02 The x the mean equal to the 9 ,100, the sigma will be the 700 divided square into the end will be 105 if we compute it we should get equal to the 907 times square with the 105 equal to the 1 .317 and then the next time we need to find the b value and the b value for the 2 -telf test equal to the 2 times the probability of the z greater than the 1 .317.

01:36 And if we use the z table, we will see this one will be 1 .317.

01:44 Then we get equal to the 2 times 0 .0 .939.

01:51 And then we have 2 times 0 .0 939, equal to the 0 .1878.

01:59 And this b value here is greater than the 10%.

02:03 So therefore it's greater than the zubon 1...

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