Question

A university is interested in evaluating registration processes. Students can register for classes by using either a telephone registration system or an online system that is accessed through the university's web site. Independent random samples of 80 students who registered by phone and 60 students who registered online were selected. Of those who registered by phone, 57 reported that they were satisfied with the registration process. Of those who registered online, 50 reported that they were satisfied. Based on these data, is it reasonable to conclude that the proportion who are satisfied is higher for those who register online? Test the appropriate hypotheses using \(\alpha=.05\).

Short answer

Yes. It is reasonable to conclude, based on the given data and testing at the 0.05 significance level, that the proportion of students who are satisfied is higher for those who register online compared to those who register by phone.

Step-by-step solution

Setting up Hypotheses

In this task, null hypothesis \(H_0\) and alternative hypothesis \(H_1\) need to be established. The null hypothesis is that the proportions are equal. Alternative hypothesis is that the proportion of satisfied students from the online registration process is higher than the phone registration process i.e. \(H_0: p_{phone} = p_{online}\), \(H_1: p_{online} > p_{phone}\)

Performing the Test

First, calculate the sample proportions, \(\hat{p}_{phone} = 57/80 = 0.7125\) and \(\hat{p}_{online} = 50/60 = 0.8333\). Calculate the pooled sample proportion considering both samples: \(p = (x_{1} + x_{2}) / (n_{1} + n_{2}) = (57 + 50) / (80 + 60) = 0.7583\). The test statistic is given by \(Z = \(\frac{(\hat{p}_{online} - \hat{p}_{phone}) - 0}{\sqrt{p(1-p)(1/n_{1}+1/n_{2})}\). In this Z test the denominator is the standard error. On substituting the values, \(Z = \(\frac{(0.8333 - 0.7125) - 0}{\sqrt{0.7583(1-0.7583)(1/80+1/60)}\ = 1.682\)

Confirm the Result

Under application of the standard normal distribution, compute the P-value corresponding to the calculated Z value. For a Z value of 1.682, the P-value is 0.0462. Compare this P-value with the given significance level \(\alpha = 0.05\). If P-value ≤ \(\alpha\), reject the null hypothesis.

Draw Conclusion

Since P-value (0.0462) < \(\alpha\) (0.05), null hypothesis that the satisfaction rates of both registration processes are equal is rejected. It can be concluded that the proportion of satisfied students in the online registration process is significantly higher than in the phone registration process.

Key concepts

Null and Alternative Hypotheses

In the realm of hypothesis testing in statistics, two fundamental hypotheses are set up to investigate the relationship between different data groups. The null hypothesis (\(H_0\)) represents the default position or status quo, suggesting that there is no significant effect or difference between the groups being compared. For our scenario involving university registration satisfaction, the null hypothesis posits that the satisfaction rates for students who registered by phone and online are equal.
The alternative hypothesis (\(H_1\text{ or }H_a\)), on the other hand, indicates the presence of an effect or a specific difference that the researcher is testing for. In this case, the alternative hypothesis suggests that the proportion of satisfied students who registered online is higher than those who registered by phone. Establishing these hypotheses is the cornerstone of hypothesis testing and sets the stage for further statistical analysis.
Formulating these hypotheses correctly is crucial as it defines the direction of the statistical test. An alternative hypothesis can be one-sided (as in this case, where we are only interested if one proportion is greater than the other) or two-sided if we are testing for any difference in either direction.

Pooled Sample Proportion

The pooled sample proportion is an estimate of the overall proportion that combines information from two independent samples when the null hypothesis assumes that the true proportions are equal. It balances the two sample proportions based on their sample sizes, providing a common ground for comparison.
In the provided university example, we calculate the pooled sample proportion (\(p\text{ pooled }\text{ or simply }p\)) by taking the total number of satisfied students from both groups and dividing by the total number of students. This accounts for the different sample sizes of the groups and hence represents the best estimate of the true population proportion under the assumption enforced by the null hypothesis.
Mathematically, it is represented as:
\[\begin{equation}p_{\text{pooled}} = \frac{x_{1} + x_{2}}{n_{1} + n_{2}}\end{equation}\]
where

• \(x_{1}\) and \(x_{2}\) are the numbers of successes in sample 1 and 2 respectively,
• \(n_{1}\) and \(n_{2}\) are the sizes of sample 1 and 2 respectively.

The pooled sample proportion is utilized in calculating the test statistic for evaluating differences in proportions.

Z Test for Proportion Difference

The Z test for proportion difference is a statistical method used to determine whether there is a significant difference between the proportions of two independent groups. It falls under the umbrella of hypothesis testing geared towards proportion comparisons.
In our exercise, we use this test to assess the difference in satisfaction rates between the two student groups who registered through different methods. The standard Z test formula is applied:
\[\begin{equation}Z = \frac{(\hat{p}_{\text{online}} - \hat{p}_{\text{phone}}) - 0}{\sqrt{p(1-p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}\end{equation}\]
where

• \(\hat{p}_{\text{online}}\) and \(\hat{p}_{\text{phone}}\) are the observed sample proportions,
• \(p\) is the pooled sample proportion assuming the null hypothesis is true,
• \(n_{1}\) and \(n_{2}\) are the respective sample sizes.

The calculated Z score or Z value is then used in conjunction with the standard normal distribution to derive the P-value, which ultimately helps in making decisions about the hypotheses.

P-value Significance Testing

The concept of P-value significance testing is fundamental to interpreting the results of most hypothesis tests, including the Z test for proportion differences. The P-value is a probability that quantifies the strength of the evidence against the null hypothesis provided by the sample data.
A lower P-value indicates stronger evidence in favor of the alternative hypothesis. This is because it tells us how likely or unlikely the observed data would be assuming the null hypothesis is true. In hypothesis testing, a predetermined significance level (\(\alpha\)) acts as a threshold to decide if the null hypothesis can be rejected. Commonly used \(\alpha\) levels are 0.05, 0.01, or 0.10, which indicate a 5%, 1%, or 10% chance of rejecting the null hypothesis when it's actually true (Type I error).
In our example, with a P-value of 0.0462 and an \(\alpha\) of 0.05, we conclude that there is sufficient evidence to reject the null hypothesis. This suggests that the data provide enough evidence, at the 5% significance level, to support the claim that satisfaction rates are higher for online registrations compared to phone registrations.