Essential Statistics
Essential Statistics
2nd Edition
ISBN: 9781259570643
Author: Navidi
Publisher: MCG
Question
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Chapter 8.2, Problem 63E

a.

To determine

State the null and alternate hypotheses.

a.

Expert Solution
Check Mark

Answer to Problem 63E

Null hypothesis: H0:μ=225.

Alternate hypothesis: H1:μ225.

Explanation of Solution

Calculation:

The output obtained using MINITAB for the hypothesis test of a population mean is given.

From the output, the given information is sample size n=50, sample mean x¯=235.32, standard error of mean SE(x¯)=4.9497, confidence interval (CI)95% CI=(225.619,245.021), test statistic value z=2.085, P-value P=0.037.

Denote μ as the true population mean.

The given test hypotheses are:

Null hypothesis:

 H0:μ=225.

That is, the true population mean is 225.

Alternate hypothesis:

H1:μ225.

That is, the true population mean is different from 225.

b.

To determine

Find the value of test statistic.

b.

Expert Solution
Check Mark

Answer to Problem 63E

The value of test statistic z is 2.085.

Explanation of Solution

Calculation:

Test statistic:

The z-test statistic is:

z=x¯μσn,

Where, x¯ be the sample mean, μ be the hypothesized mean and σ be the standard deviation and n be the sample size.

From the given output, the value of test statistic z is 2.085.

c.

To determine

Find the P-value.

c.

Expert Solution
Check Mark

Answer to Problem 63E

The P-value is 0.037.

Explanation of Solution

Calculation:

P-value:

P-value is defined as the probability of obtaining a result at least as extreme as the observed value, given that the null hypothesis, H0 is true. The test is said to be statistically significant, when P-value<α, for level of significance α.

From the given output, the P-value is 0.037.

d.

To determine

Decide whether the null hypothesis H0 is rejected at α=0.05 level.

d.

Expert Solution
Check Mark

Answer to Problem 63E

The null hypothesis H0 is rejected at α=0.05 level.

Explanation of Solution

It is given that the significance level α=0.05.

From previous part (a), the P-value is 0.037.

Decision based on the P-value method level of significance α:

  • If Pvalueα, reject H0.
  • If Pvalue>α, fail to reject H0.

Conclusion:

The significance level is, α=0.05 and the P-value is 0.037.

Here, the P-value of 0.037 is less than the significance level 0.05.

That is, P-value(=0.037)<α(=0.05).

Therefore, the null hypothesis is rejected.

e.

To determine

Decide whether the null hypothesis H0 is rejected at α=0.01 level.

e.

Expert Solution
Check Mark

Answer to Problem 63E

The null hypothesis H0 is not rejected at α=0.01 level.

Explanation of Solution

It is given that the significance level α=0.01.

From previous part (a), the P-value is 0.037.

Decision based on the P-value method level of significance α:

  • If Pvalueα, reject H0.
  • If Pvalue>α, fail to reject H0.

Conclusion:

The significance level is, α=0.01 and the P-value is 0.037.

Here, the P-value of 0.037 is greater than the significance level 0.01.

That is, P-value(=0.037)>α(=0.01).

Therefore, the null hypothesis is not rejected.

f.

To determine

Find the value of test statistic.

f.

Expert Solution
Check Mark

Answer to Problem 63E

The value of test statistic is 1.07.

Explanation of Solution

Calculation:

The given test hypotheses are:

Null hypothesis:

 H0:μ=230.

That is, the true population mean is 230.

Alternate hypothesis:

H1:μ>230.

That is, the true population mean is greater than 230.

Test statistic:

The z-test statistic is:

z=x¯μσn,

Where, x¯ be the sample mean, μ be the hypothesized mean and σ be the standard deviation and n be the sample size.

Software procedure:

Step by step procedure to find the test statistic using the MINITAB software:

  • Choose Stat > Basic Statistics > 1-Sample Z.
  • In Summarized data select the Sample size as 50, Sample mean as 235.32.
  • Select Known standard deviation as 35.
  • Enter Hypothesized mean as 230.
  • Select Options.
  • Choose Mean>hypothesized mean in Alternative hypothesis.
  • Click OK in all dialogue boxes.

The output using Minitab is given below:

Essential Statistics, Chapter 8.2, Problem 63E

From the MINITAB output, the test statistic, that is, the z-value is 1.07.

Thus, the value of test statistic z is 1.07.

g.

To determine

Identify the P-value.

g.

Expert Solution
Check Mark

Answer to Problem 63E

The P-value is 0.1412.

Explanation of Solution

From the previous part (f), it is observed from the MINITAB output, the P-value is 0.1412.

Thus, the P-value is 0.1412.

h.

To determine

Decide whether the null hypothesis H0 in the part (f) is rejected at α=0.05 level.

h.

Expert Solution
Check Mark

Answer to Problem 63E

The null hypothesis H0 is not rejected at α=0.05 level.

Explanation of Solution

It is given that the significance level α=0.05.

From previous part (a), the P-value is 0.1412.

Decision based on the P-value method level of significance α:

  • If Pvalueα, reject H0.
  • If Pvalue>α, fail to reject H0.

Conclusion:

The significance level is, α=0.05 and the P-value is 0.037.

Here, the P-value of 0.1412 is greater than the significance level 0.05.

That is, P-value(=0.1412)>α(=0.05).

Therefore, the null hypothesis is not rejected.

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Chapter 8 Solutions

Essential Statistics

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