Aresearcher wants to prove that there is a difference in the average life spans between men and women in japan. let mu1 = average life span of japanese women and mu2 = average life span of japanese men. a random sample of 10 women showed an average lifespan of 83 years, with a sample standard deviation of 7 years. a random sample of 10 men showed an average lifespan of 77 years, with a sample standard deviation of 6.4 years. assume that life spans are normally distributed and that the population variances are equal. if alpha = .05 and the null hypothesis is mu1 – mu2 = 0, what is (are) the critical value(s) for the hypothesis test?

t < -2.101, and t > 2.101

Step-by-step explanation:

We are running a hypothesis for the difference of 2 sample means, assuming normality, and assuming that population variances are equal.  This determines which test we run.  

We have:

Sample 1: (women)

n = 10

x = 83

s = 7

Sample 2: (men)

n = 10

x = 77

s = 6.4

The hypothesis for the test are:

H0:  µ1 - µ2 = 0

Ha:  µ1 - µ2 ≠ 0

The significance level is 5%.  The degrees of freedom is 2 less than the sum of the sample size, in this case, 18.  Our t-value is:  2.101

Our critical values for the test statistic are:  t < -2.101, and t > 2.101

Area is 91.5ft squared

Step-by-step explanation:

6 * 14 = 84

The rectangle is 84ft.

7 + 4 = 11

14 - 11 = 3

5 * 3 = 15

15 / 2 = 7.5

84 + 7.5 = 91.5

Area is 91.5ft squared

hey steph