You wish to test the following claim (H.) at a significance level of a = 0.02. H¸:µ = 62.8 Ha:µ # 62.8 You believe the population is normally distributed and you know the standard deviation is o = 13.4. You obtain a sample mean of M = 61 for a sample of size n = 44.

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You wish to test the following claim (\(H_a\)) at a significance level of \(\alpha = 0.02\).

\[
H_o: \mu = 62.8
\]
\[
H_a: \mu \neq 62.8
\]

You believe the population is normally distributed and you know the standard deviation is \(\sigma = 13.4\). You obtain a sample mean of \(M = 61\) for a sample of size \(n = 44\).

What is the test statistic for this sample? (Report answer accurate to three decimal places.)

\[ \text{test statistic} = \_\_\_ \]

What is the p-value for this sample? (Report answer accurate to four decimal places.)

\[ \text{p-value} = \_\_\_ \]

The p-value is...

- [ ] less than (or equal to) \(\alpha\)
- [ ] greater than \(\alpha\)
Transcribed Image Text:You wish to test the following claim (\(H_a\)) at a significance level of \(\alpha = 0.02\). \[ H_o: \mu = 62.8 \] \[ H_a: \mu \neq 62.8 \] You believe the population is normally distributed and you know the standard deviation is \(\sigma = 13.4\). You obtain a sample mean of \(M = 61\) for a sample of size \(n = 44\). What is the test statistic for this sample? (Report answer accurate to three decimal places.) \[ \text{test statistic} = \_\_\_ \] What is the p-value for this sample? (Report answer accurate to four decimal places.) \[ \text{p-value} = \_\_\_ \] The p-value is... - [ ] less than (or equal to) \(\alpha\) - [ ] greater than \(\alpha\)
Expert Solution
Step 1

The test statistic is,

z=M-μσn=61-62.813.444=-1.82.0201=-0.891

Thus, the test statistic is -0.891.

The hypothesis is two tail. The p-value for two tailed test is,

p-value=Pz<-0.891  or  z>0.891=2×Pz<-0.891

The probability of z less than –0.891 can be obtained using the excel formula “=NORM.S.DIST(–0.891,TRUE)”. The probability value is 0.18647.

The required P-value is,

p-value=2×Pz<-0.891=2×0.18647=0.3729

Thus, the p-value is 0.3729.

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