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Chapter 9: Q.95 (page 592)

Sampling shoppers A marketing consultant observes 50 consecutive shoppers at a supermarket, recording how much each shopper spends in the store. Explain why it would not be wise to use these data to carry out a significance test about the mean amount spent by all shoppers at this supermarket.

Short Answer

Expert verified

Convenience sample.

Step by step solution

01

Step-1 Given Information

Given in the question that, A marketing consultant observes 50consecutive shoppers at a supermarket, we have to Explain why it would not be wise to use these data to carry out a significance test about the mean amount spent by all shoppers at this supermarket.

02

Step-2 Explanation

A convenient sample is a non-probability sampling method in which the sample is taken from a group of people who are easy to touch or encounter. A convenience survey, for example, would entail standing outside a mall or a grocery store and asking customers to complete questionnaires. This kind of sampling is also known as grab sampling or availability sampling. The poll is a choice study, as the shoppers were chosen because they were "conveniently" consecutive.

There would also be shoppers who went shopping on the same day of the week and at the same time, implying that a segment of the population was left out.

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Most popular questions from this chapter

Asked to explain the meaning of 鈥渟tatistically significant at the A 0.05 level,鈥� a

student says, 鈥淭his means that the probability that the null hypothesis is true is less than 0.05.鈥� Is this explanation correct? Why or why not?

In planning a study of the birth weights of babies whose mothers did not see a

doctor before delivery, a researcher states the hypotheses as

H0:渭<1000gramsHa:渭=900grams

A random sample of 100likely voters in a small city produced 59voters in favor of Candidate A. The observed value of the test statistic for testing the null hypothesis H0:p=0.5versus the alternative hypothesis Ha:p=0.5is

(a)z=0.59-0.50.59(0.41)100

(b). z=0.59-0.50.5(0.5)100

(c). z=0.5-0.590.59(0.41)100

(d). z=0.5-0.590.5(0.5)100

(e).t=0.59-0.50.5(0.5)100

Study more! The significance test in Exercise 76yields a P-value of 0.0622.

(a) Describe a Type I and a Type II error in this setting. Which type of error could you have made in Exercise 76? Why?

(b) Which of the following changes would give the test a higher power to detect 渭=120minutes: using 伪=0.01or 伪=0.10? Explain.

Is your food safe? Do you feel confident or not confident that the food available at most grocery stores is safe to eat? when a gallup poll asked this question, 87%of the sample said they were confident." Callup announced the poll's margin of error for 95%confidence as卤3 percentage points. Which of the following sources of error are included in this margin of error? Explain.

(a) Gallup dialed landline telephone numbers at random and so missed all people without landline phones, including people whose only phone is a cell phone.

(b) Some people whose numbers were chosen never answered the phone in several calls or answered but refused to participate in the poll.

(c) There is chance variation in the random selection of telephone numbers.
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