Given the following data, test the hypothesis that the average body weight is th

Question

Given the following data, test the hypothesis that the average body weight is the same for two independent diagnosis groups from one hospital: State all the elements of your test. Take alpha = .05. Find the exact p - value of your test, draw your conclusion based on the p-value. State the assumptions needed to carry on the test in a)? Refer to Problem 6, above. Test the hypothesis that the populations corresponding to the two groups have equal variances against a two sided alternative. Use a = 0.05, give all the elements of your test. Find the exact p-value and write your conclusion based on the p-value.

Explanation / Answer

Question 6. a)

Here, we have to use the two sample t test for the population means. The null and alternative hypothesis for this test is given as below:

Null hypothesis: H0: The average body weight is same for two independent diagnosis groups.

Alternative hypothesis: Ha: The average body weight is not same for two independent diagnosis groups.

We are given alpha = 0.05

This is a two tailed test.

The test statistic formula is given as below:

Test statistic = t = (X1bar – X2bar) / sqrt[(S1^2/N1)+(S2^2/N2)]

From the given data, we have

X1bar = 140.4285714

S1 = 15.4365

N1 = 7

X2bar = 143

S2 = 16.6012

N2 = 6

Test statistic = t = (140.4286 – 143) / sqrt[(15.4365^2/7)+(16.6012^2/6)]

Test statistic = t = -0.2875

Degrees of freedom = 10 (approx.)

Critical value = -2.2281 and 2.2281

Test statistic lies between two critical values.

So, do not reject the null hypothesis that the average body weight is same for two independent diagnosis groups.

Part b

P-value =0.7796 by using t-table

Alpha value = 0.05

P-value > Alpha value

So, we do not reject the null hypothesis that the average body weight is same for two independent diagnosis groups.

Part c

For this test we assume the assumption of normality and we also assume that the given two samples are independent from each other.

Question 7

Part a

Here, we have to test the same claim by assuming equal population variances which is given as below:

Test statistic = t = (140.4286 – 143) / sqrt[((1/7)+(1/6))(15.4365^2)+(16.6012^2)]

Test statistic = t = -0.2893

Degrees of freedom = 7 + 6 – 2 = 13 – 2 = 11

Critical values = -2.2010 and 2.2010

Test statistic value lies between critical values.

So, we do not reject the null hypothesis that the average body weight is same for two independent diagnosis groups.

Part b

P-value = 0.7777 by using t-table

Alpha value = 0.05

P-value > alpha value

So, we do not reject the null hypothesis that the average body weight is same for two independent diagnosis groups.

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