Question

You wish to test the following claim (H1H1) at a significance level of α=0.01α=0.01.       Ho:μ=82.5Ho:μ=82.5       H1:μ≠82.5H1:μ≠82.5...

You wish to test the following claim (H1H1) at a significance level of α=0.01α=0.01.

      Ho:μ=82.5Ho:μ=82.5
      H1:μ≠82.5H1:μ≠82.5

You believe the population is normally distributed, but you do not know the standard deviation. You obtain a sample of size n=449n=449 with mean M=81.7M=81.7 and a standard deviation of SD=10.7SD=10.7.

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 82.5.
  • There is not sufficient evidence to warrant rejection of the claim that the population mean is not equal to 82.5.
  • The sample data support the claim that the population mean is not equal to 82.5.
  • There is not sufficient sample evidence to support the claim that the population mean is not equal to 82.5.

Homework Answers

Answer #1

Solution :

The null and alternative hypothesis are

H0 : = 82.5 ......... Null hypothesis

Ha :    82.5 .......... Alternative hypothesis

Here, n = 449, = 81.7, s = 10.7

The test statistic t is,

t =  

= [81.7 - 82.5]/[10.7 /449]

= -1.584

The value of the test​ statistic t = -1.584

Now ,

d.f. = n - 1 = 449 - 1 = 448

sign in Ha indicates that the test is TWO TAILED.

t = -1.584

So , using calculator ,

p value = 0.1138

Since,

p value is greater than the significance level 0.01 .

Decision: Fail to Reject the null hypothesis H0

Conclusion :

There is not sufficient evidence to warrant rejection of the claim that the population mean is not equal to 82.5.

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