Chapter 8.1, Problem 9UYK

(a)

To determine

To test: The significance test on the competitor’s product which provided better UVA and UVB protection.

Answer to Problem 9UYK

Solution: The results are summarized as $\widehat{p}=0.35$, $z=-1.34$, and $p-\text{value}=0.1802$.

Explanation of Solution

Calculation: The provided data shows that $n=20$ and $X=7$. The parameter p is the proportion of people who received better UVA and UVB protection from the product. If there is no difference between the two lotions, then $p=0.5$. The hypothesis is given by:

$\begin{array}{c}{H}_0:p=0.5\\ H_a:p\ne 0.5\end{array}$

Here, $H_0$ is the null hypothesis and $H_a$ is the alternative hypothesis.

The expected number of successes and failures are $20\times 0.5=10$ and $20\times 0.5=10$. As both the expected numbers are at least 10, use z-test to compute the test statistic.

The formula of test statistic for z-test is defined as:

$z=\frac{\widehat{p}-p_0}{\sqrt{\frac{p_0\left(1-p_0\right)}{n}}}$

Here, $\widehat{p}$ is the sample proportion and $p_0$ is the hypothesized value of the population proportion.

The formula for sample proportion $\widehat{p}$ is defined as:

$\widehat{p}=\frac{X}{n}$

Here,

$\begin{array}{c}X=\text{number of succeses in the sample}\\ n=\text{sample size}\end{array}$

Substitute $X=7$ and $n=20$ in the above defined formula to get the required sample proportion. So,

$\begin{array}{c}\widehat{p}=\frac{X}{n}\\ =\frac{7}{20}\\ =0.35\end{array}$

Therefore, the sample proportion $\widehat{p}$ is obtained as 0.35. Substitute the obtained values of $\widehat{p}=0.35$ and $p_0=0.5$ in the formula of test statistic. So, the test statistic is calculated as:

$\begin{array}{c}z=\frac{\widehat{p}-p_0}{\sqrt{\frac{p_0\left(1-p_0\right)}{n}}}\\ =\frac{0.35-0.5}{\sqrt{\frac{0.5\left(1-0.5\right)}{20}}}\\ =\frac{-0.15}{0.1118}\\ =-1.34\end{array}$

Use Table A of standard normal probabilities to obtained the probability.

The probability is obtained as $P\left(Z<-1.34\right)=0.0901$.

Therefore, the p-value, which is the area in both the tails, is obtained as:

$\begin{array}{c}p-\text{value}=2\times 0.0901\\ =0.1802\end{array}$

Conclusion: Since $p-\text{value}=0.1802$ is larger than 0.05, the null hypothesis is not rejected. Conclude that the sunblock testing data is not significant. The provided data does not support the proposed advertising claim.

To determine

To find: The comparison of results obtained in previous part with the results obtained in Example 8.5.

Answer to Problem 9UYK

Solution: The results obtained in the previous part are the same with the results obtained in Example 8.5.

Explanation of Solution

Calculation: The results obtained in the previous part are summarized as:

$\widehat{p}=0.35,z=-1.34,\text{ and }p-\text{value}=0.1802$

The results obtained in Example 8.5 are summarized as:

$\widehat{p}=0.65,z=1.34,\text{ and }p-\text{value}=0.1802$

Though the value of the sample proportion and the test statistic differ, the p-value is same for both the cases. Therefore, the conclusion for both the cases is the same, that is, the sunblock testing data is not significant.

(b)

To determine

To find: The 95% confidence interval for the hypothesis that the competitor’s product provides better protection.

Answer to Problem 9UYK

Solution: The 95% confidence interval for the hypothesis that the competitor’s product provides better protection is $\underset{\_}{\left(14.1\%,55.9\%\right)}$.

Explanation of Solution

Calculation: The formula for confidence interval in the arrangement of estimate plus or minus the margin of error is:

$\widehat{p}\pm m$

Here, m is the margin of error, which is defined as:

$m=z^{*}\times S{E}_{\widehat{p}}$

Here, $z^{*}$ is the critical value of the standard normal density curve.

The sample proportion is obtained as 0.35 in the previous part. Substitute this proportion in the standard error formula. So,

$\begin{array}{c}S{E}_{\widehat{p}}=\sqrt{\frac{\widehat{p}\left(1-\widehat{p}\right)}{n}}\\ =\sqrt{\frac{0.35\left(1-0.35\right)}{20}}\\ =\sqrt{\frac{0.2275}{20}}\\ =0.1067\end{array}$

Therefore, the standard error is obtained as 0.2067. The value of $z^{*}$ for 95% confidence level is $z^{*}=1.96$, which is obtained from the standard normal table.

So, the margin of error is obtained as:

$\begin{array}{c}m=z^{*}\times S{E}_{\widehat{p}}\\ =1.96\times 0.1067\\ =0.209132\end{array}$

Substitute the values of margin of error and sample proportion in the formula for confidence interval of the arrangement of estimate, plus or minus the margin of error. Therefore, the confidence interval is obtained as:

$\begin{array}{c}\widehat{p}\pm m=0.35\pm 0.209132\\ =\left(0.35-0.209132,0.35+0.209132\right)\\ =\left(0.140868,0.559132\right)\\ =\left(14.1\%,55.9\%\right)\end{array}$

To determine

To explain: The comparison of 95% confidence interval for the competitor’s product providing better protection and the 95% confidence interval for your product providing better protection.

Answer to Problem 9UYK

Solution: The widths of the confidence intervals obtained in both the cases are the same.

Explanation of Solution

The 95% confidence interval for your product providing better protection against the competitor product is obtained as $\left(44.10\%,85.90\%\right)$ in example 8.5. This shows that the length of the confidence interval is $\left(85.9-44.1=41.8\right)$.

The 95% confidence interval for the competitor’s product providing better protection is obtained as $\left(14.1\%,55.9\%\right)$ in the previous part. This shows that the length of the confidence interval is $\left(55.9-14.1=41.8\right)$.

Therefore, these results show that the width of the confidence intervals is the same but only the lower and the upper limit changes.