Question

A newly hired telemarketer is told he will probably make a sale on about \(12 \%\) of his phone calls. The first week he called 200 people, but only made 10 sales. Should he suspect he was misled about the true success rate? Explain.

Short answer

Yes, the telemarketer should suspect a different success rate, as the results are significantly below expected.

Step-by-step solution

Define the null and alternative hypotheses

The null hypothesis, denoted as \(H_0\), assumes that the telemarketer's success rate matches the claimed rate of \(12\%\) (or 0.12). The alternative hypothesis, \(H_a\), assumes that the success rate is different from \(12\%\). Thus, \(H_0: p = 0.12\) and \(H_a: p eq 0.12\), where \(p\) is the true success rate.

Calculate the expected number of sales

If the telemarketer calls 200 people, the expected number of sales can be calculated using the formula for expected value: \(E = n \times p\), where \(n = 200\) and \(p = 0.12\). Hence, \(E = 200 \times 0.12 = 24\) sales.

Calculate the standard error

The standard error (SE) for the proportion is given by the formula \(SE = \sqrt{\frac{p(1-p)}{n}}\). Substituting the known values, \(SE = \sqrt{\frac{0.12 \times 0.88}{200}} = \sqrt{\frac{0.1056}{200}} \approx 0.023\).

Calculate the z-score

The z-score measures how far the observed number of sales (10) is from the expected number (24), in terms of standard deviations. It's calculated as \(z = \frac{\text{observed} - \text{expected}}{SE}\). Substituting the values, \(z = \frac{10 - 24}{0.023} \approx -14.35\).

Interpret the z-score

A typical z-score for hypothesis testing falls within the range of -1.96 to 1.96 for a 95% confidence interval. The z-score of approximately -14.35 is far below -1.96, which indicates that the observed result (10 sales) is significantly lower than expected, assuming the success rate is truly \(12\%\).

Conclusion

Since the z-score is far outside the typical range, we reject the null hypothesis. This means there is statistical evidence to suggest that the true success rate is different from the given \(12\%\).

Key concepts

Null Hypothesis

The null hypothesis is a foundational concept in hypothesis testing. It is essentially a statement we aim to test and possibly disprove. In this scenario, the null hypothesis, denoted as \(H_0\), is that the telemarketer's true success rate matches the claimed rate of 12%, or 0.12 in decimal form. This means we initially assume that everything is as expected and that any deviations are due to random chance.

In contrast, the alternative hypothesis, \(H_a\), suggests the opposite - that the success rate is different from 0.12. This is critical because hypothesis testing is all about weighing evidence to either support or refute our assumptions.

• If the evidence strongly contradicts \(H_0\), we "reject" it, implying that the alternative may be true.
• If the evidence is not strong enough, we "fail to reject" \(H_0\), which means it remains a plausible explanation.

Standard Error

The standard error (SE) provides insight into the variability of a sample statistic. In the context of our problem, it measures how much the observed proportion of calls that led to sales is expected to vary from the true proportion of 12%.

It's crucial because it helps us determine the precision of our estimate of the telemarketer’s success rate. The formula for SE of a proportion is given by \(SE = \sqrt{\frac{p(1-p)}{n}}\), where \(p\) is the assumed success rate and \(n\) is the sample size. For this problem, we calculated it as approximately 0.023, which tells us that the observed proportion's deviations from the true proportion should generally fall within this range.

A smaller standard error implies a less spread-out dataset, whereas a larger one indicates more variability around the true value.

Z-score

The z-score is a statistic that quantifies how far away our observed results are from what's expected under the null hypothesis, in units of standard deviations. For the telemarketer, a z-score was calculated to see how unusual his 10 sales were compared to the expected 24 sales.

The formula for a z-score is \(z = \frac{\text{observed} - \text{expected}}{SE}\). Here, the observed is 10, the expected is 24, and the SE is 0.023. We found a z-score of approximately -14.35. The negative number indicates that the observed value is less than expected.

Such an extreme z-score suggests that the observed result of 10 sales is incredibly far from what we'd predict if the success rate were truly 12%. In hypothesis testing, a common range for z-scores is -1.96 to 1.96, corresponding to a 95% confidence level. If our z-score falls outside this range, it's an indication to reconsider our assumptions, as in this case.

Confidence Interval

A confidence interval represents the range within which we expect the true parameter, like a success rate, to lie with a certain level of confidence. Although the exercise doesn't directly calculate a confidence interval, knowing about it aids in understanding hypothesis test conclusions.

For a 95% confidence interval, we would use our sample data to estimate the range where we might find the true success rate. For a normal distribution, a 95% confidence interval would generally be computed as the observed proportion ± 1.96 times the standard error.

This interval helps us understand the reliability of our estimation. If a claimed rate, like 12% in our problem, lies outside the computed interval, it suggests that the claim might not be valid. Thus, a confidence interval provides additional insight into the validity of our hypothesis test outcomes.