Question

In: Statistics and Probability

The table below lists the weights of some colleage students in September and later in February...

The table below lists the weights of some colleage students in September and later in February of their freshman year.

September weight 62 58 71 67 51 60 72 50 66 57 70 82 62 56 73 51 71 55 70 82 56 60 69 58
February weight 68 56 67 72 56 58 77 55 69 56 65 76 57 59 77 56 68 56 64 76 56 58 75 54

At the 0.05 significance level, test the claim of no difference between September weights and February weights.
The test statistic is________
The critical value is ________

Solutions

Expert Solution

Solution:

Here, we have to use paired t test.

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

Null hypothesis: H0: There is no difference between September weights and February weights.

Alternative hypothesis: Ha: There is a difference between September weights and February weights.

H0: µd = 0 versus Ha: µd ≠ 0

This is a two tailed test.

Test statistic for paired t test is given as below:

t = (Dbar - µd)/[Sd/sqrt(n)]

From given data, we have

Dbar = -0.0833

Sd = 4.3928

n = 24

df = n – 1 = 23

α = 0.05

t = (Dbar - µd)/[Sd/sqrt(n)]

t = (-0.0833 – 0)/[ 4.3928/sqrt(24)]

t = -0.0929

Test statistic = t = -0.0929

The test statistic is -0.0929.
The critical value is ±2.0687

(by using t-table)

The p-value by using t-table is given as below:

P-value = 0.9268

P-value > α = 0.05

So, we do not reject the null hypothesis

There is sufficient evidence to conclude that there is no difference between September weights and February weights.


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