Fast Food Dinner Service Times Data Set 25 “Fast Food” in Appendix B lists drivethrough service times (seconds) for dinners at McDonald’s, Burger King, and Wendy’s. Using those times with a TI-83>84 Plus calculator yields the following display. Using a 0.05 significance level, test the claim that the three samples are from populations with the same mean. What do you conclude?

Short Answer

Expert verified

The null hypothesis will be rejected at a 0.05 significance level. There is not enough evidence to support the claim that the three samples are from populations with the same mean.

It can be concluded that three samples are observed from populations that have different mean times for service.

Step by step solution

01

Given information

Drive-through service times for dinners at three places are compared.

The level of significance is 0.05.

02

Identify the hypotheses as per the claim

Let\[{\mu _1},{\mu _2},{\mu _3}\]be the actual mean service times for McDonald’s, Burger King, and Wendy’s, respectively.

The hypotheses are stated below.

\(\begin{array}{l}{H_0}:{\mu _1} = {\mu _2} = {\mu _3}\\{H_a}:\;{\rm{at}}\;{\rm{least}}\;{\rm{one}}\;{\mu _i}\;{\rm{is}}\;{\rm{different}}{\rm{.}}\end{array}\)

03

Identify the p-value from the output

Decision rule:

  • For a p-value less than 0.05, the null hypothesis is rejected.
  • For a p-value greater than 0.05, the null hypothesis is failed to be rejected.

From the output of the TI-83/84 Plus calculator, the p-value is marked as\(P = 0.0045462237\). It is lesser than 0.05. Thus, the null is rejected at the 0.05 significance level.

04

Interpret the result

Rejecting the null hypothesis implies that the mean service time for at least one of the three dinners is statistically different from the others. As a result, it can be concluded that the populations from which three samples have been selected have varying mean time.

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Most popular questions from this chapter

Weights The weights (kg) in the following table are from Data Set 1 “Body Data” in Appendix B. Results from two-way analysis of variance are also shown. Use the displayed results and use a 0.05 significance level. What do you conclude?


Female

Male

18-29

63.4

57.8

52.6

46.9

61.7

61.5

77.2

50.4

97

76.1

71.6

64.9

144.9

96.4

80.7

84.4

63.9

79

99.4

64.1

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110.5

84.6

133.3

90.2

125.7

105.3

115.5

75.3

92.8

57.7

96.2

56.4

107.4

99.5

64.8

94.7

74.2

112.8

72.6

91.4

50-80

103.2

48.3

87.8

101.3

67.8

45.2

79.8

60.1

68.5

43.3

84.8

127.5

89.9

75.3

110.2

72.3

77.2

86.5

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73.1

Cola Weights Identify the value of the test statistic in the display included with Exercise 1. In general, do larger test statistics result in larger P-values, smaller P-values, or P-values that are unrelated to the value of the test statistic

In Exercises 1–4, use the following listed arrival delay times (minutes) for American Airline flights from New York to Los Angeles. Negative values correspond to flights that arrived early. Also shown are the SPSS results for analysis of variance. Assume that we plan to use a 0.05 significance level to test the claim that the different flights have the same mean arrival delay time.

Flight 1

-32

-25

-26

-6

5

-15

-17

-36

Flight 19

-5

-32

-13

-9

-19

49

-30

-23

Flight 21

-23

28

103

-19

-5

-46

13

-3

Why Not Test Two at a Time? Refer to the sample data given in Exercise 1. If we want to test for equality of the three means, why don’t we use three separate hypothesis tests for\({\mu _1} = {\mu _2},{\mu _1} = {\mu _3}\;and\;{\mu _2} = {\mu _3}\)?

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