Question

You wish to test the following claim (H1H1) at a significance level of α=0.002α=0.002.

      Ho:μ=89.3Ho:μ=89.3
      H1:μ≠89.3H1:μ≠89.3

You believe the population is normally distributed and you know the standard deviation is σ=13.2σ=13.2. You obtain a sample mean of ¯x=94x¯=94 for a sample of size n=75n=75.

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

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

The p-value is...

• less than (or equal to) αα
• greater than αα

This test statistic leads to a decision to...

• reject the null
• accept the null
• fail to reject the null

As such, the final conclusion is that...

• There is sufficient evidence to warrant rejection of the claim that the population mean is not equal to 89.3.
• There is not sufficient evidence to warrant rejection of the claim that the population mean is not equal to 89.3.
• The sample data support the claim that the population mean is not equal to 89.3.
• There is not sufficient sample evidence to support the claim that the population mean is not equal to 89.3.

Answer 1

Given that, sample size (n) = 75, sample mean = 94 and

population standard deviation = 13.2

The null and alternative hypotheses are,

H0 : μ = 89.3

H1 : μ ≠ 89.3 (claim)

This hypothesis test is a two-tailed test.

Test statistic is,

=> Test statistic = 3.084

Using Excel we find, the p-value

Excel Command : = 2 * (1 - NORMSDIST(3.084)) = 0.0020

=> P-value = 0.0020

The P-value is less than (or equal to) α

This test statistic leads to a decision to reject the null.

As such, the final conclusion is that,

There is sufficient evidence to warrant rejection of the claim that the population mean is not equal to 89.3.