You wish to test the following claim (HaHa) at a significance level of =0.05=0.0

Question

You wish to test the following claim (HaHa) at a significance level of =0.05=0.05.  dd denotes the mean of the difference between pre-test and post-test scores.       

Ho:d=0Ho:d=0
Ha:d<0Ha:d<0

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

Make sure to choose the appropriate column for Sample 1 and 2 so that you are estimating the improved post-test score: 212-1

What is the test statistic for this sample?

test statistic =  Round to 3 decimal places.

What is the p-value for this sample? Round to 4 decimal places.

p-value =  

The p-value is...

less than (or equal to)

greater than

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 mean difference of post-test from pre-test is less than 0.

There is not sufficient evidence to warrant rejection of the claim that the mean difference of post-test from pre-test is less than 0.

The sample data support the claim that the mean difference of post-test from pre-test is less than 0.

There is not sufficient sample evidence to support the claim that the mean difference of post-test from pre-test is less than 0.

post-test pre-test 56.7 61.5 60.5 57.7 70.6 94.6 55 23.6 53.6 58.4 60 52.3 56.1 69.1 59.3 53.3 64.4 41.8 57.9 23.5 59.8 44.8 60.4 67.2 62.2 81.1 62.1 74.1 63.8 44.2 64 60.7 61.1 80 56.2 13.7 55.6 53.3 52 29.9 62.7 63.9

Explanation / Answer

HEre we will use paired data t - test

Ho:d=0
Ha:d<0

The difference table with mean and standard deviation of difference

Average difference dbar = -5.0143

Stanard deviation of difference sd = 18.5370

number of paired data n = 21

Standard error of paired data difference sed = sd/ sqrt (n) = 18.5370/sqrt(21) = 4.0451

Test statistic

t = dbar/sed = -5.0143/4.0451 = -1.2396

p - value = 0.1147

The p - value is greater than alpha

This leads that we failed to reject the null hypothesis.

Final conclusion is that There is not sufficient sample evidence to support the claim that the mean difference of post-test from pre-test is less than 0..OPtion D is correct.

post-test pre-test Difference 56.7 61.5 4.8 60.5 57.7 -2.8 70.6 94.6 24 55 23.6 -31.4 53.6 58.4 4.8 60 52.3 -7.7 56.1 69.1 13 59.3 53.3 -6 64.4 41.8 -22.6 57.9 23.5 -34.4 59.8 44.8 -15 60.4 67.2 6.8 62.2 81.1 18.9 62.1 74.1 12 63.8 44.2 -19.6 64 60.7 -3.3 61.1 80 18.9 56.2 13.7 -42.5 55.6 53.3 -2.3 52 29.9 -22.1 62.7 63.9 1.2 Average -5.0143 Std. Dev. 18.5370

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