Research claims that all men drive at most the speed of women. to test a counter

Question

Research claims that all men drive at most the speed of women. to test a counter claim tha the average speed at which all men drive is higher than that of women a reseracher recorded the speeds of vehicles, where a sample of 27 cars driven by men had a mean speed of 72.0 mph with a standard deviation of of 2.2 mph, and a sample of 18 cars driven by women had a mean speed of 70.0 mph with a standard devaition of 2.5 mph. the data is normally distributed.

For the appropriate hypothesis test, the null and alternative hypotheses are what?

the obtained value of the appropriate hypothesis test statisticis what?

comparing the obtained and critical values(a=.05), is the researchers counter claim supported?

for the difference between the populations attributes, the 95% confidence interval is?

Explanation / Answer

Solution:-

The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. We work through those steps below:

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

Null hypothesis: 1 - 2 = 0 , i.e., the men and women drive at same speed.
Alternative hypothesis: 1 - 2 0, i.e., the men drive at higher speed than women.

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 0.50. 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 = sqrt[(2.22/27) + (2.52/18] = 0.4714

DF = (s12/n1 + s22/n2)2 / { [ (s12 / n1)2 / (n1 - 1) ] + [ (s22 / n2)2 / (n2 - 1) ] }
DF = (2.22/27 + 2.52/18)2 / { [ (2.22 / 18)2 / (26) ] + [ (2.52 / 18)2 / (17) ] }
DF = 0.27718 / (0.00278 + 0.00709) = 28.083
t = [ (x1 - x2) - d ] / SE = [ (72 - 70) - 0 ] / 0.4714 = 4.24268

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 is more extreme than 4.24268; that is, less than 4.24268 or greater than 4.24268.

We use the t Distribution Calculator to find P(t < 4.24268)

The P-Value is 0.000217.

The result is significant at p < 0.05.

Interpret results. Since the P-value (0.000217) is less than the significance level (0.50), we cannot accept the null hypothesis.

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