Arbitron Media Research Inc. conducted a study of the iPod listening habits of men and women. One facet of the study involved the mean listening time. It was discovered that the mean listening time for a sample of 13 men was 35 minutes per day. The standard deviation was 8 minutes per day. The mean listening time for a sample of 11 women was also 35 minutes, but the standard deviation of the sample was 18 minutes. Use a two-tailed test and at 0.10 significance level, can we conclude that there is a difference in the variation in the listening times for men and women?

Since the critical f-value of the test statistic is less than the f value of 2.9130, we will fail to reject the null hypothesis and conclude that there's no sufficient evidence to support the claim that there is a difference in the variation in the listening times for men and women

Step-by-step explanation:

We are given;

Sample size for men; n1 = 13

Sample size for women; n2 = 11

standard deviation for men; s1 = 8 minutes

Standard deviation for women; s2 = 18 minutes.

Significance level; α = 0.1

Let's state the hypothesis;

Null hypothesis;H0: (μ1)² = (μ2)²

Alternative hypothesis;Ha: (μ1)² ≠ (μ2)²

The value of the test statistic would be;

F = (s1)²/(s2)²

F = 8²/18² = 0.1975

Now, degree of freedom for n1 is;

DF1 = n1 - 1

DF1 = 13 - 1

DF1 = 12

Also, degree of freedom for n2 is;

DF2 = 11 - 1

DF2 = 10

Now, since it's two tailed, we will make use of α/2 for the F-distribution table.

Thus, α/2 = 0.1/2 = 0.05

So,from the f-table attached, at df1 = 12 and df2 = 10,the F-Critical value is;

F_α/2 = 2.9130

Since,the critical f-value of the test statistic is less than 2.9130, we will fail to reject the null hypothesis and conclude that there's no sufficient evidence to support the claim that there is a difference in the variation in the listening times for men and women

he spend 5.1 minutes chatting with his friends

step-by-step explanation:

2/5+1/2=9/10

6/1-9/10=5 1/10

597

step-by-step explanation: