Question

a. The average expenditure per student (based on average daily attendance) for a certain school year was $10,337 with a population standard deviation of $1560. A survey for the next school year of 150 randomly selected students resulted in a sample mean of $10, 798. Find the P-value? Should the null hypothesis be rejected at alpha = .05 level of significance? b. The average expenditure per student (based on average daily attendance) for a certain school year was $10,337 with a population standard deviation of $1560. A survey for the next school year of 150 randomly selected students resulted in a sample mean of $10, 798. Find the 95% confidence level? Should the null hypothesis be rejected? c. The average expenditure per student (based on average daily attendance) for a certain school year was $10,337 with a population standard deviation of $1560. A survey for the next school year of 150 randomly selected students resulted in a sample mean of $10, 798. Find the Z statistics? Should the null hypothesis be rejected at alpha = .05 level of significance?

          a. The average expenditure per student (based on average daily
attendance) for a certain school year was $10,337 with a population
standard deviation of $1560. A survey for the next school year of
150 randomly selected students resulted in a sample mean of $10,
798. Find the P-value? Should the null hypothesis be rejected at
alpha = .05 level of significance?
b. The average expenditure per student (based on average
daily attendance) for a certain school year was $10,337 with a
population standard deviation of $1560. A survey for the next
school year of 150 randomly selected students resulted in a sample
mean of $10, 798. Find the 95% confidence level? Should the null
hypothesis be rejected?
c. The average expenditure per student (based on average
daily attendance) for a certain school year was $10,337 with a
population standard deviation of $1560. A survey for the next
school year of 150 randomly selected students resulted in a sample
mean of $10, 798. Find the Z statistics? Should the null hypothesis
be rejected at alpha = .05 level of significance?

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Added by Mark A.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

Instant Answer

Solved by Expert Madhur L

03/30/2023

Step 1

Define the null hypothesis and alternative hypothesis. Null hypothesis (H0): The average expenditure per student remains the same, i.e., μ = $10,337. Alternative hypothesis (H1): The average expenditure per student has changed, i.e., μ ≠ $10,337. Show more…

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a. The average expenditure per student (based on average daily attendance) for a certain school year was $10,337 with a population standard deviation of $1560. A survey for the next school year of 150 randomly selected students resulted in a sample mean of $10, 798. Find the P-value? Should the null hypothesis be rejected at alpha = .05 level of significance? b. The average expenditure per student (based on average daily attendance) for a certain school year was $10,337 with a population standard deviation of $1560. A survey for the next school year of 150 randomly selected students resulted in a sample mean of $10, 798. Find the 95% confidence level? Should the null hypothesis be rejected? c. The average expenditure per student (based on average daily attendance) for a certain school year was $10,337 with a population standard deviation of $1560. A survey for the next school year of 150 randomly selected students resulted in a sample mean of $10, 798. Find the Z statistics? Should the null hypothesis be rejected at alpha = .05 level of significance?

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Transcript

00:01 Hello students here for the question number a they given that mu is equal to 10 237 and the x bar is equal to 10798 sigma is equal to 1560 then n is equal to 150 then the given level of significance alpha is 0 .05 now we want to frame the hypothesis mu is equal to 10337 and on the alternative hypothesis h -alpha mu not equal to 10 -337.

00:40 The formula for calculating the test statistic is that x bar minus mu divided by sigma by square root of n.

00:47 Now substitute the known values.

00:49 We get 10798 minus 10337 divided by 1560 divided by 1560 divided by square root of 150.

00:59 We get our z value as 3 .619.

01:03 Then here the p value for this test is 0 .003, which is less than the alpha 0 .05.

01:14 So here we reject the null hypothesis h0.

01:27 Then for the subpart b, we've given that x par is equal to 10798 and the sigma is 1560, is equal to 150 then the given confidence interval is 95 percentage here we need to find the confidence interval the formula for calculating the confidence interval is x bar plus r minus z alpha by 2 into sigma by square root of n now substitute the known values here we get z alpha by 2 which is z 0 .05 by 2 we get 1 .96 that is 10798 plus or minus 1 .96 into 1560 divided by square root of 150.

02:25 By simplifying this we get 10798 plus or minus 249 .6474.

02:34 Therefore here we get 10 -548 .35 and 11047 .65.

02:44 To two decimal places here we get.

02:48 Here the given mu is 10 -337...

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