Question

A machine shop that manufactures toggle levers has both a day and a night shift. A toggle lever is defective if a standard nut cannot be screwed onto the threads. Let \(p_{1}\) and \(p_{2}\) be the proportion of defective levers among those manufactured by the day and night shifts, respectively. We shall test the null hypothesis, \(H_{0}: p_{1}=p_{2}\), against a two-sided alternative hypothesis based on two random samples, each of 1000 levers taken from the production of the respective shifts. Use the test statistic \(Z^{*}\) given in Example \(6.5 .3\). (a) Sketch a standard normal pdf illustrating the critical region having \(\alpha=0.05\). (b) If \(y_{1}=37\) and \(y_{2}=53\) defectives were observed for the day and night shifts, respectively, calculate the value of the test statistic and the approximate \(p-\) value (note that this is a two-sided test). Locate the calculated test statistic on your figure in Part (a) and state your conclusion. Obtain the approximate \(p\) -value of the test.

Short answer

Test statistic Z: -1.63, p-value: 0.1032. Since the p-value is greater than the significance level of 0.05, we do not reject the null hypothesis. Hence, there is no significant difference in the proportion of defective levers between the day shift and the night shift.

Step-by-step solution

State The Hypotheses

The null hypothesis is that the two proportions are equal, that is, \(H_{0}: p_{1} = p_{2}\). The alternative hypothesis is that the two proportions are not equal, which can be written as \(H_{a}: p_{1} \neq p_{2}\).

Formulate An Analysis Plan

The plan is to use a Z test for two proportions. For this, we first calculate the pooled proportion \(p\), which is the total number of success outcomes divided by the total number of trials, using the formula \(p = (y_{1}+y_{2})/(n_{1}+n_{2})\). We determine the standard deviation using \(se = \sqrt{p*(1-p)*[(1/n_{1}) + (1/n_{2})]}\), and use this to calculate the test statistic using \(Z = (p_{1} - p_{2})/se\). Finally, we calculate the p-value in a two-sided test using the standard deviation from the Z distribution.

Analyze Sample Data

Using the data provided, we can calculate the proportions \(p_{1}= y_{1}/n_{1} = 37/1000 = 0.037\) and \(p_{2} = y_{2}/n_{2} = 53/1000 = 0.053\). The pooled proportion is \(p = (y_{1}+y_{2})/(n_{1}+n_{2}) = (37+53)/(1000+1000) = 90/2000 = 0.045\). The standard error is \(se = \sqrt{p*(1-p)*[(1/n_{1}) + (1/n_{2})]} = \sqrt{0.045*(1-0.045)*[(1/1000) + (1/1000)]} = 0.00983\). Then, the Z statistic value is \((p_{1} - p_{2})/se = (0.037 - 0.053)/0.00983 = -1.63\).

Interpret The Results

Now we will find the p-value. In a two-tailed test, we multiply the one-tailed probability by 2. Using a Z-table or a statistical software, we conclude that the one-tailed p-value is 0.0516. Therefore, the two-tailed p-value equals 0.0516 * 2 = 0.1032. As the p-value is greater than the significance level of 0.05, we do not reject the null hypothesis. Hence, there is not enough evidence at the 5% significance level to conclude that the proportion of defective levers is different between the day and night shifts.

Key concepts

Null Hypothesis

The null hypothesis, often symbolized as \(H_0\), is a statement used in statistical analysis that proposes no significant difference or effect exists between certain specified populations. In the context of hypothesis testing, it acts as the default assumption that the observed data is not driven by any underlying effect.

In the exercise, the null hypothesis is stating that the proportion of defective levers produced by the day shift \(p_1\) is equal to the proportion produced by the night shift \(p_2\), meaning \(H_0: p_1 = p_2\).

The null hypothesis is critical because it provides a baseline or a point of comparison for testing if the data provides enough evidence to support an alternative hypothesis, which in our exercise is \(H_a: p_1 eq p_2\). Without a clearly defined null hypothesis, the process of scientific inquiry through hypothesis testing could lack direction and clarity.

Z Test for Proportions

The Z test for proportions is a statistical method used to determine if there is a difference between sample proportions. It is particularly useful when dealing with large sample sizes.

In this exercise, we need to compare the proportions of defective levers from two different shifts. We start by calculating a pooled proportion \(p\), which is the combined proportion of defectives from both shifts. To find \(p\), we use the formula:

• \(p = \frac{y_1 + y_2}{n_1 + n_2}\)

where \(y_1\) and \(y_2\) are the number of defectives from the day and night shifts, respectively, and \(n_1\) and \(n_2\) are the sample sizes.

We then calculate the standard error (\(se\)) to measure the variability in sample proportions:

• \(se = \sqrt{p(1-p) \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}\)

Finally, the Z statistic is determined by:

• \(Z = \frac{p_1 - p_2}{se}\)

where \(p_1\) and \(p_2\) are the observed sample proportions. The value of Z tells us how far, in standard deviations, the observed difference is from the expected difference (which is zero under \(H_0\)).

Two-sided Test

A two-sided test, also known as a two-tailed test, is used in hypothesis testing when we want to determine if there is a significant difference in either direction away from zero. This means that deviations in both directions are considered significant.

In the context of our exercise, the alternative hypothesis \(H_a: p_1 eq p_2\) is two-sided. This is because we are interested in knowing whether the proportions of defective levers are different, regardless of which shift has a higher or lower proportion.

The critical region for a two-sided test is found on both tails of the normal distribution. When evaluating the Z statistic, if it lies in either tail beyond the critical value, we may reject the null hypothesis.

The decision to use a two-sided test reflects our lack of prior assumption about which proportion might be greater, emphasizing the neutrality in the hypothesis testing approach.

P-value

The p-value is a probability measure that helps us determine the strength of evidence against the null hypothesis. It is the probability of obtaining a result at least as extreme as the one observed, assuming the null hypothesis is true.

In simpler terms, the p-value tells us how likely it is to get our sample data by random chance. A small p-value indicates strong evidence against the null hypothesis, leading us to consider rejecting it. Conversely, a large p-value suggests weak evidence against the null hypothesis.

In our case, after calculating the Z statistic of -1.63, we seek the p-value for this test. Since our test is two-sided, we double the one-tailed p-value. The resultant two-tailed p-value of 0.1032 is above the common significance level of \(\alpha = 0.05\). Hence, we do not reject the null hypothesis, indicating there's not enough evidence at the 5% significance level to claim a difference in defect proportions between the shifts.