You wish to test the following claim (HaHa) at a significance level of =0.05=0.0

Question

You wish to test the following claim (HaHa) at a significance level of =0.05=0.05.

      Ho:=85.9Ho:=85.9
      Ha:85.9Ha:85.9

You believe the population is normally distributed, but you do not know the standard deviation. You obtain a sample of size n=25n=25with mean M=87.3M=87.3 and a standard deviation of SD=5.4SD=5.4.

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)

greater than

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 85.9.

There is not sufficient evidence to warrant rejection of the claim that the population mean is not equal to 85.9.

The sample data support the claim that the population mean is not equal to 85.9.

There is not sufficient sample evidence to support the claim that the population mean is not equal to 85.9.

Explanation / Answer

Solution:

Here, we have to use one sample t test for the population mean.

The null and alternative hypotheses for this test are given as below:

Null hypothesis: H0: µ = 85.9 versus Alternative hypothesis: Ha: µ 85.9

This is a two tailed test.

The test statistic formula is given as below:

Test statistic = t = (Xbar - µ) / [S/sqrt(n)]

Where µ is population mean, Xbar is sample mean, S is sample standard deviation and n is the sample size.

We are given

Xbar = 87.3

µ = 85.9

S = 5.4

n = 25

Also, we are given

Level of significance or alpha value = = 0.05

Now, degrees of freedom = n – 1 = 25 – 1 = 24

Lower critical value = -2.0639 (by using t-table or excel)

Upper critical value = 2.0639 (by using t-table or excel)

Now, we have to find test statistic value.

Test statistic = t = (87.3 – 85.9) / [5.4/sqrt(25)]

Test statistic = t = 1.4/[5.4/5] = 1.4/1.08 = 1.296296

Test statistic = t = 1.296296

P-value = 0.2072

= 0.05

P-value >

So, we do not reject the null hypothesis

There is not sufficient sample evidence to support the claim that the population mean is not equal to 85.9.

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