Each person in a random sample of 221 male teenagers and a random sample of 301

Question

Each person in a random sample of 221 male teenagers and a random sample of 301 female teenagers was asked how many hours he or she spent online in a typical week. The sample mean and standard deviation were 15.2 hours and 11.3 hours for males and 14.3 and 11.8 for females. (Use a statistical computer package to calculate the P-value. Use males females. Round your test statistic to two decimal places, your df down to the nearest whole number, and your P-value to three decimal places.) t = df = P =

Each person in a random sample of 221 male teenagers and a random sample of 301 female teenagers was asked how many hours he or she spent online in a typical week. The sample mean and standard deviation were 15.2 hours and 11.3 hours for males and 14.3 and 11.8 for females. (Use a statistical computer package to calculate the P-value. Use males females. Round your test statistic to two decimal places, your df down to the nearest whole number, and your P-value to three decimal places.) t = df = P =

Each person in a random sample of 221 male teenagers and a random sample of 301 female teenagers was asked how many hours he or she spent online in a typical week. The sample mean and standard deviation were 15.2 hours and 11.3 hours for males and 14.3 and 11.8 for females. (Use a statistical computer package to calculate the P-value. Use males females. Round your test statistic to two decimal places, your df down to the nearest whole number, and your P-value to three decimal places.) t = df = P =

Explanation / Answer

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: 1 = 2
Alternative hypothesis: 1 2

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the difference between sample means is too big or if it is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a two-sample t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = sqrt[(s12/n1) + (s22/n2)]
SE = 1.01998
DF = 520
t = [ (x1 - x2) - d ] / SE

t = 0.88

where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is the size of sample 1, n2 is the size of sample 2, x1 is the mean of sample 1, x2 is the mean of sample 2, d is the hypothesized difference between the population means, and SE is the standard error.

Since we have a two-tailed test, the P-value is the probability that a t statistic having 520 degrees of freedom is more extreme than 0.88; that is, less than -0.88 or greater than 0.88.

Thus, the P-value = 0.3793

Interpret results. Since the P-value (0.3793) is greater than the significance level (0.05), we have to accept the null hypothesis.

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