OPTIMA Statistics

A pharmaceutical company has developed a new drug for the common cold. They claim that this drug shortens the duration of the cold in 80% of individuals. A physician believes that this proportion is inflated. He finds 115 individuals who have just developed a cold, administers the drugs, and observes that 89 of them recover more quickly than is typical. Find the null and alternative hypotheses for this test.

A government official claims that 50% of cars on the road have under-inflated tires, and are therefore getting less-than-optimal gas mileage. To test this claim, a random sample of 600 cars is selected and 279 of them are found to have under-inflated tires. Find the null and alternative hypotheses for this test.

A scout leader claims that being a boyscout promotes character development and respect for the law. In fact, he claims that fewer than 10% of boys who spend at least one year in a boyscout troupe will get into trouble with the law before their 20th birthday. To test this claim, you take a random sample of 80 names from the rosters of boyscout troupes fifteen years in the past and find that 11 of these former boy-scouts have had trouble with the law. Find the null and alternative hypotheses for this test.

A pharmaceutical company has developed a new drug for the common cold. They claim that this drug shortens the duration of the cold in 80% of individuals. A physician believes that this proportion is inflated. He finds 115 individuals who have just developed a cold, administers the drugs, and observes that 89 of them recover more quickly than is typical. Find the test statistic for this sample.

A government official claims that 50% of cars on the road have under-inflated tires, and are therefore getting less-than-optimal gas mileage. To test this claim, a random sample of 600 cars is selected and 279 of them are found to have under-inflated tires. Find the test statistic for this sample.

A scout leader claims that being a boyscout promotes character development and respect for the law. In fact, he claims that fewer than 10% of boys who spend at least one year in a boyscout troupe will get into trouble with the law before their 20th birthday. To test this claim, you take a random sample of 80 names from the rosters of boy-scout troupes fifteen years in the past and find that 11 of these former boyscouts have had trouble with the law. Find the test statistic for this sample.

A pharmaceutical company has developed a new drug for the common cold. They claim that this drug shortens the duration of the cold in 80% of individuals. A physician believes that this proportion is inflated. He finds 115 individuals who have just developed a cold, administers the drugs, and observes that 89 of them recover more quickly than is typical. Conduct a traditional hypothesis test at the \(\alpha = 0.05\) significance level.

Solution

Recall that the null and alternative hypothesis were:

\begin{align*} H_0\amp:\ p \geq 0.80\\ H_A\amp:\ p \lt 0.80 \end{align*}

We computed the test statistic as follows:

\begin{equation*} z_{\text{test}} = \frac{0.7739 - 0.80}{\sqrt{\frac{(0.8)(0.2)}{115}}} \approx -0.70\text{.} \end{equation*}

Now we must identify the rejection region and the critical value. Since the alternative hypothesis involves \(\lt\text{,}\) this is a left-tailed test with the entire significance level of \(\alpha = 0.05\) in that left tail. This gives a critical value \(z_\alpha = -1.645\) as shown below.

Because our test statistic of \(-0.70\) is not less than the critical value \(-1.645\text{,}\) it is not in the rejection region. We therefore fail to reject the null hypothesis. There is no statistically significant evidence that the pharmaceutical company is inflating the claims about this drug's effectiveness.

A government official claims that 50% of cars on the road have under-inflated tires, and are therefore getting less-than-optimal gas mileage. To test this claim, a random sample of 600 cars is selected and 279 of them are found to have under-inflated tires. Use a traditional hypothesis test at the \(\alpha = 0.10\) significance level to test the government officials claim.

Solution

The hypotheses for this test were:

\begin{align*} H_0\amp:\ p = 0.50\\ H_A\amp:\ p \not= 0.50 \end{align*}

And we found that the test statistic was:

\begin{equation*} z_{\text{test}} = \frac{0.465- 0.50}{\sqrt{\frac{(0.5)(0.5)}{600}}} \approx - 1.71\text{.} \end{equation*}

The rejection region in this two-tailed test is show below, separated from the body of the standard normal distribution by the critical values \(-1.645\) and \(+1.645\text{.}\)

Since our test statistic of \(z_\text{test} = -1.71\) is in one of these rejection regions, we must reject the null hypothesis. There is evidence tending towards significance that the true proportion of cars on the road with under-inflated tires is different from 0.50.

A scout leader claims that being a boyscout promotes character development and respect for the law. In fact, he claims that fewer than 10% of boys who spend at least one year in a boyscout troupe will get into trouble with the law before their 20th birthday. To test this claim, you take a random sample of 80 names from the rosters of boy-scout troupes fifteen years in the past and find that 11 of these former boyscouts have had trouble with the law. Conduct a traditional hypothesis test to test this claim at the \(\alpha = 0.10\) significance level.

Solution

Recall that the hypotheses are:

\begin{align*} H_0\amp:\ p \geq 0.10\\ H_A\amp:\ p \lt 0.10 \end{align*}

The test statistic under the null hypothesis above is:

\begin{equation*} z_{\text{test}} = \frac{0.1375 - 0.1000}{\sqrt{\frac{(0.1)(0.9}{80}}} \approx 1.12\text{.} \end{equation*}

Because the alternative hypothesis involves \(\lt\text{,}\) this is a left-tailed test with the entire \(\alpha = 0.10\) in the left tail. The corresponding critical value is -1.28 as shown below.

Our test statistic of 1.12 is clearly not in the rejection region. Therefore we fail to reject the null hypothesis. There is not even evidence tending towards significance that fewer than 10% of boyscout members have trouble with the law before their 20th birthday. For more examples of conducting a traditional hypothesis test, see the following videos.

A pharmaceutical company has developed a new drug for the common cold. They claim that this drug shortens the duration of the cold in 80% of individuals. A physician believes that this proportion is inflated. He finds 115 individuals who have just developed a cold, administers the drugs, and observes that 89 of them recover more quickly than is typical. Conduct a p-value hypothesis test at the \(\alpha = 0.05\) significance level.

Solution

Recall that the null and alternative hypothesis were:

\begin{align*} H_0\amp:\ p \geq 0.80\\ H_A\amp:\ p \lt 0.80 \end{align*}

We computed the test statistic as follows:

\begin{equation*} z_{\text{test}} = \frac{0.7739 - 0.80}{\sqrt{\frac{(0.8)(0.2)}{115}}} \approx -0.70\text{.} \end{equation*}

The p-value for our test statistic in this left-tailed test (because the alternative uses \(\lt\)) is the probability of getting a test statistic further into that left tail than -0.70.

As depicted above, this is

\begin{equation*} P(Z \lt -0.70) = 0.2420\text{.} \end{equation*}

Clearly our p-value is greater than the significance level of \(\alpha = 0.05\) (and in fact would be greater than any commonly used significance level). We therefore fail to reject the null hypothesis. There is no evidence (statistically significant or otherwise) that the pharmaceutical company is inflating the claims about this drug's effectiveness.

A government official claims that 50% of cars on the road have under-inflated tires, and are therefore getting less-than-optimal gas mileage. To test this claim, a random sample of 600 cars is selected and 279 of them are found to have under-inflated tires. Use a p-value hypothesis test at the \(\alpha = 0.10\) significance level to test the government officials claim.

Solution

The hypotheses for this test were:

\begin{align*} H_0\amp:\ p = 0.50\\ H_A\amp:\ p \not= 0.50 \end{align*}

And we found that the test statistic was:

\begin{equation*} z_{\text{test}} = \frac{0.465- 0.50}{\sqrt{\frac{(0.5)(0.5)}{600}}} \approx - 1.71\text{.} \end{equation*}

Since this is a two-tailed test, the p-value of our test statistic will be twice the probability of being further into the left tail (since the test statistic is negative).

As shown above, this gives

\begin{equation*} 2 \times P(Z \lt -1.71) = 2(0.0436) = 0.0872\text{.} \end{equation*}

Because this p-value is less than the significance level we reject the null hypothesis. There is evidence tending towards significance that the true proportion of cars on the road with under-inflated tires is different from 0.50.

A scout leader claims that being a boyscout promotes character development and respect for the law. In fact, he claims that fewer than 10% of boys who spend at least one year in a boyscout troupe will get into trouble with the law before their 20th birthday. To test this claim, you take a random sample of 80 names from the rosters of boyscout troupes fifteen years in the past and find that 11 of these former boyscouts have had trouble with the law. Conduct a p-value hypothesis test of this claim at the \(\alpha = 0.01\) significance level.

Solution

Recall that the hypotheses are:

\begin{align*} H_0\amp:\ p \geq 0.10\\ H_A\amp:\ p \lt 0.10 \end{align*}

The test statistic under the null hypothesis above is:

\begin{equation*} z_{\text{test}} = \frac{0.1375 - 0.1000}{\sqrt{\frac{(0.1)(0.9)}{80}}} \approx 1.12\text{.} \end{equation*}

Now this is a left-tailed test because the alternative hypothesis involves \(\lt\text{.}\) Therefore, the p-value of our test statistic is the probability of being further into the left tail than 1.12. Note, however, that 1.12 is on the right of the mean of 0 as shown below. All this means is that not only is there no evidence tosupport the claim, but the evidence seems to point to the opposite being true—that is to more than 10% being in trouble with the law.

Finishing this test, the p-value is \(P(Z \lt 1.12) = 0.8686\text{.}\) This is clearly much larger than the significance level of \(\alpha = 0.01\text{.}\) We therefore fail to reject the null hypothesis. There is not even evidence tending towards significance that fewer than 10% of boyscout members have trouble with the law before their 20th birthday.

A pharmaceutical company has developed a new drug for the common cold. They claim that this drug shortens the duration of the cold in 80% of individuals. A physician believes that this proportion is inflated. He finds 115 individuals who have just developed a cold, administers the drugs, and observes that 89 of them recover more quickly than is typical. If you conduct a hypothesis test at the \(\alpha = 0.05\) significance level using this sample, and in actuality the drug is only effective in 75% of patients, what error will be made?

A government official claims that 50% of cars on the road have under-inflated tires, and are therefore getting less-than-optimal gas mileage. To test this claim, a random sample of 600 cars is selected and 279 of them are found to have under-inflated tires. If a hypothesis test is conducted at the \(\alpha = 0.10\) significance level to test the government officials claim, and the true proportion of cars with under-inflated tires actually is 0.50, what type of error has been made?