Question

I am using a course that requires me to use excel.I am able to figure everything out except the test statistic

A marine biologist claims that the mean length of mature female pink seaperch is different in fall and winter. A sample of 15 mature female pink seaperch collected in fall has a mean length of 108 millimeters and a standard deviation of 15 millimeters. A sample of 8 mature female pink seaperch collected in winter has a mean length of 106 millimeters and a standard deviation of 13 millimeters. At alphaequals0.02​, can you support the marine​ biologist's claim? Assume the population variances are equal. Assume the samples are random and​ independent, and the populations are normally distributed. Complete parts​ (a) through​ (e) below.

Answer 1

i am using excel to solve the problem.

i am denoting mean length of pink seaperch in fall as sample 1 and mean length of pink seaperch in winter as sample 2.  

steps:-

open an excel spreadsheet add -ins Phstat two sample tests (summarized data)pooled-variance t test in hypothesized difference type 0 , in level of signifiacnce type 0.02 in population 1 sample , type 15 in sample size, 108 in sample mean and 15 in sample standard deviation in population 2 sample , type 8 in sample size, 106 in sample mean and 13 in sample standard deviation in test options select two- tail test ok

your excel output be:-

Pooled-Variance t Test for the Difference Between Two Means
(assumes equal population variances)
Data
Hypothesized Difference	0
Level of Significance	0.02
Population 1 Sample
Sample Size	15
Sample Mean	108
Sample Standard Deviation	15
Population 2 Sample
Sample Size	8
Sample Mean	106
Sample Standard Deviation	13
Intermediate Calculations
Population 1 Sample Degrees of Freedom	14
Population 2 Sample Degrees of Freedom	7
Total Degrees of Freedom	21
Pooled Variance	206.3333
Standard Error	6.2887
Difference in Sample Means	2.0000
t Test Statistic	0.3180
Two-Tail Test
Lower Critical Value	-2.5176
Upper Critical Value	2.5176
p-Value	0.7536
Do not reject the null hypothesis

ANSWERS:-

a). hypothesis:-

b).test statistic = 0.3180

c). p value = 0.7536

d).decision:-

p value = 0.7536 > 0.02 (alpha)

so, we fail to reject the null hypothesis.

e).conclusion:-

there is not sufficient evidence to support the claim that the mean length of mature female pink seaperch is different in fall and winter at 0.02 level of significance.

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