Question

the following data for problems 15-17 Current Product 30 24.5 5.5 Difference 30 2.1 5.05 New Produdct x (in hours) s (in hours 30 26.6 4.75 15. A company wants to compare its new antiperspirant with the leading antiperspirant currently on the market for preventing wetness and odor. The currently selling clinical strength antiperspirant claims to keep you dry and odor free for 24 hours. The new antiperspirant contains a novel ingredient, which scientists believe will outlast the performance of the current product. You design an experiment to compare the two products by randomly assigning one of the two products to 60 college athletes. That is, 30 college athletes are randomly selected to receive the Current Product and the remaining 30 college athletes receive the New Product. Each person records the length of time (in hours) between the time that they applied the product to the first time that they experience wetness or odor. Does the data support the current antiperspirant's claim? Construct a 95% confidence interval and interpret results. Is the new antiperspirant more effective in keeping athletes drier than the current antiperspirant? Perform a 5-step test of hypothesis. Use b. a-0.05

Answer 1

Solution:-

15)

a) 95% confidence interval for the difference in means is C.I = (- 4.7563, 0.5563).

$C.I = \left (\bar{x}_{1}-\bar{x}_{2} \right )\pm t_{\alpha /2}\times \sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}}$

C.I = (24.5 - 26.6) + 2.002*1.3268

C.I = -2.1 + 2.6563

C.I = (- 4.7563, 0.5563)

Since the confidence interval contains 0, hence we do not have sufficient evidence in the favor of the claim that new antiperspirant is more effective in keeping athletes drier than the current antiperspirant.

b)

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

Null hypothesis: u₁> u₂
Alternative hypothesis: u₁ < u₂

Note that these hypotheses constitute a one-tailed test.

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 = 1.3268
DF = 58

t = [ (x₁ - x₂) - d ] / SE

t = - 1.583

where s₁ is the standard deviation of sample 1, s₂ is the standard deviation of sample 2, n₁ is thesize 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 population means, and SE is the standard error.

The observed difference in sample means produced a t statistic of -1.583

Therefore, the P-value in this analysis is 0.059.

Interpret results. Since the P-value (0.059) is greater than the significance level (0.05), we cannot reject the null hypothesis.

From the above test we do not have sufficient evidence in the favor of the claim that new antiperspirant is more effective in keeping athletes drier than the current antiperspirant.