Question

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

      Ho:μ=87.9Ho:μ=87.9
      Ha:μ≠87.9Ha:μ≠87.9

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

Column A	Column B	Column C	Column D	Column E
94.5	85.8	82.7	93.3	86.4
81.6	96	92.2	82.5	92.5
83.8	93.3	79.2	85.4	100
93.9	79.2	82.9	84	87.9
86.7	93.6	87.3	89.9	91
70.7	91.3	86	102.4	81.1
86.9	91.7	86.2	91.3	84
99.2	76.2	96	72.1	91
92.2	88.5	93	81.1	85.4
81.6	90.4	83.2	94.9	100
76.7	80.6	81.1	93.6	82.9

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 87.9.
• There is not sufficient evidence to warrant rejection of the claim that the population mean is not equal to 87.9.
• The sample data support the claim that the population mean is not equal to 87.9.
• There is not sufficient sample evidence to support the claim that the population mean is not equal to 87.9.

Answer 1

Given: = 87.9

From the data: = 87.58, s = 6.9313, n = 55, = 0.05

The Hypothesis:

H0: = 87.9

Ha: 87.9

This is a 2 tailed test

__________________

The Test Statistic: Since the population standard deviation is unknown, we use the students t test.

The test statistic is given by the equation:

t observed = -0.35

The p Value:    The p value for t = , for degrees of freedom (df) = n-1 = , is; p value = 0.7277

The p value is : Option 2: Greater than

The test statistic leads to a decision of: Option 3: Fail to Reject the Null.

The Final Conclusion: Option 4: There is not sufficient evidence to support the claim that the that the population mean is not equal to 87.9.

__________________________________

Calculation for the mean and standard deviation:

Mean = Sum of observation / Total Observations

Standard deviation = SQRT(Variance)

Variance = Sum Of Squares (SS) / n - 1, where SS = SUM(X - Mean)².

Sno	X	Mean	(X - Mean)2
1	94.5	87.58	47.8864
2	81.6	87.58	35.7604
3	83.8	87.58	14.2884
4	93.9	87.58	39.9424
5	86.7	87.58	0.7744
6	70.7	87.58	284.9344
7	86.9	87.58	0.4624
8	99.2	87.58	135.0244
9	92.2	87.58	21.3444
10	81.6	87.58	35.7604
11	76.7	87.58	118.3744
12	85.8	87.58	3.1684
13	96	87.58	70.8964
14	93.3	87.58	32.7184
15	79.2	87.58	70.2244
16	93.6	87.58	36.2404
17	91.3	87.58	13.8384
18	91.7	87.58	16.9744
19	76.2	87.58	129.5044
20	88.5	87.58	0.8464
21	90.4	87.58	7.9524
22	80.6	87.58	48.7204
23	82.7	87.58	23.8144
24	92.2	87.58	21.3444
25	79.2	87.58	70.2244
26	82.9	87.58	21.9024
27	87.3	87.58	0.0784
28	86	87.58	2.4964
29	86.2	87.58	1.9044
30	96	87.58	70.8964
31	93	87.58	29.3764
32	83.2	87.58	19.1844
33	81.1	87.58	41.9904
34	93.3	87.58	32.7184
35	82.5	87.58	25.8064
36	85.4	87.58	4.7524
37	84	87.58	12.8164
38	89.9	87.58	5.3824
39	102.4	87.58	219.6324
40	91.3	87.58	13.8384
41	72.1	87.58	239.6304
42	81.1	87.58	41.9904
43	94.9	87.58	53.5824
44	93.6	87.58	36.2404
45	86.4	87.58	1.3924
46	92.5	87.58	24.2064
47	100	87.58	154.2564
48	87.9	87.58	0.1024
49	91	87.58	11.6964
50	81.1	87.58	41.9904
51	84	87.58	12.8164
52	91	87.58	11.6964
53	85.4	87.58	4.7524
54	100	87.58	154.2564
55	82.9	87.58	21.9024
Total	4816.9		2594.308

n	55
Sum	4816.9
Average	87.58
SS	2594.308
Variance = SS/n-1	48.04274
Std Dev	6.9313