Question

You wish to test the claim μ<89.8μ<89.8 at a significance level of 0.0010.001.

You believe the population is normally distributed, but you do not know the standard deviation. You obtain the following sample of data:

data
43.4
100.8
71
39.6
81.7
83.7
55.9
74
63.1
54.7
86.5

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

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

The test statistic is...

• in the critical region
• not in the critical region

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...

• I was able to show that the average score equals 89.8
• I was unable to show that the average score equals 89.8
• I was unable to show that the μ<89.8μ<89.8.
• I was able to show that the average score is less than 89.8.
• I was able to show that the average score is greater than 89.8.

Answer 1

Here claim is that μ<89.8

So hypothesis is vs

For the given data sample mean is

Create the following table.

data	data-mean	(data - mean)2
43.4	-25.1818	634.12305124
100.8	32.2182	1038.01241124
71	2.4182	5.84769124
39.6	-28.9818	839.94473124
81.7	13.1182	172.08717124
83.7	15.1182	228.55997124
55.9	-12.6818	160.82805124
74	5.4182	29.35689124
63.1	-5.4818	30.05013124
54.7	-13.8818	192.70437124
86.5	17.9182	321.06189124

So

The t-critical value for a left-tailed test, for a significance level of α=0.001 is

tc​=−4.144

Graphically

Hence test statistics is

As test statistics do not fall in the rejection region, we fail to reject the null hypothesis

Hence we do not have sufficient evidence to support the claim that μ<89.8

Correct answers are

• fail to reject the null
• I was unable to show that the μ<89.8