You wish to test the following claim ( Ha) at a significance level of α=0.02
H0:μ1=μ2
Ha:μ1≠μ2

You believe both populations are normally distributed, but you do not know the standard deviations for either. We will assume that the population variances are not equal.

You obtain a sample of size n1=16 with a mean of M1=66.2 and a standard deviation of SD1=10.6 from the first population. You obtain a sample of size n2=26 with a mean of M2=68.8 and a standard deviation of SD2=12.9 from the second population.

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? For this calculation, use the conservative under-estimate for the degrees of freedom. The degrees of freedom is the minimum of n1 - 1 and n2 - 1. (Report answer accurate to four decimal places.)

p-value =

Question

08-03-2022
Mathematics

contestada

You wish to test the following claim ( Ha) at a significance level of α=0.02
H0:μ1=μ2
Ha:μ1≠μ2

You believe both populations are normally distributed, but you do not know the standard deviations for either. We will assume that the population variances are not equal.

You obtain a sample of size n1=16 with a mean of M1=66.2 and a standard deviation of SD1=10.6 from the first population. You obtain a sample of size n2=26 with a mean of M2=68.8 and a standard deviation of SD2=12.9 from the second population.

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? For this calculation, use the conservative under-estimate for the degrees of freedom. The degrees of freedom is the minimum of n1 - 1 and n2 - 1. (Report answer accurate to four decimal places.)

p-value =

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Аноним Аноним · Answer 1 · 2022-03-08T20:54:00+01:00

Answer:

Since we assume the variances of the two populations are the same, we can use the pooled variance in the computation of the t-statistic:pooled variance (Sp^2) = (n1-1)s1^2 + (n2-1)*s2^2 / (n1 + n2 - 2) = (13-1)*19.4^2 + (19-1)*12.6^2 / (13 +19- 2) = 245.8Sp = 15.67t-calc = (mean1 - mean2)/ (Sp * sqrt(1/n1 + 1/n2)) = (83.6-72.9)/(15.67 * sqrt(1/13 + 1/19) = 1.897since this is a right tailed test, the p-value = P(t > 1.897) at n1 + n2 - 2 = 30 degrees of freedom = 0.0337so would not reject.

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