In this Lab, you will be asked to analyze completed test review questions.? It is your turn to be the evaluator.? Check each problem to see if th

Hypothesis Testing Review –

ce department’s claim

o 𝜇 150.00 𝜇 > 150.00

𝑡 = ̅

√

𝑡 = . √

, 159 2.35

p .0022

el of significance the evidence supports the police department’s

o 𝜇 = 15 𝜇 ≠ 15

𝑡 = ̅

√

𝑡 = . .

√

2.118

20 2.093

p p

ibutions from a university’s alumni and supporters is considered

o 𝑝 ≥ .15 𝑝 < .15

�̂� = �̂� = _ = .16

𝑧 = ( )

𝑧 = . .

. ( . ) 0.443

1.645

p 06710. 0.6710

o 𝜇 = 𝜇 𝜇 ≠ 𝜇

𝑡 = ( ̅ ̅ ) ( )

𝑡 = ( . ) ( )

. .

25, 29

p 0.1225 0.1225

6

4

o 𝜇 ≤ 𝜇 𝜇 > 𝜇

𝑡 = ( ̅ ̅ ) ( )

𝑧 = ( . . ) ( )

. . 0.797

5 1 1.48

p

There is not sufficient evidence to say that there is a reduction in Gary’s mean time

o

1010 980 1170 1200 1040 1280 1450 1470 1500 1100 1260 1190 1280 1170 1370 1440 1500 1520 90 280 20 80 130 90 10 30 20

o 𝜇 ≤ 60 𝜇 > 60

𝑡 = √

𝑡 = .

.

√

.780

9 2.896

p 0.2304 0.2304

o 𝑝 ≥ 𝑝 𝑝 < 𝑝

�̂� = �̂� = �̂� = = .75

�̅� = = = .74

𝑧 = ( ) ( )

̅( ̅)( ) 𝑧 =

(. . )

. ( . )( )

1.43

1.64

p 1.64

university’s claim that the retention rate improved

manager’s claim using 0.10 level of significance. The owner surveyed 739 customers over a period of time to record each customer’s preferred day of the week. Here’s what he found.

103 103 126 103 111 96 97

o 𝑝 = 𝑝 = 𝑝 = 𝑝 = 𝑝 = 𝑝 = 𝑝

o 2 (𝒐 − 𝒆)𝟐

𝒆 103 2.57 6.6122 .0626 103 105.71 2.57 6.6122 .0626 126 105.71 20.429 417.33 3.953 103 105.71 2.57 6.6122 .0626 111 105.71 5.4286 29.469 .27914 96 105.71 9.571 91.612 .86777 97 105.71 9.571 73.469 .69592

� 5.984

7 – –

2 10.645 From x2

p

At the .10 level of significance the evidence does not support the manager’s claim

15 29 94 102 105 301 30 44 99 97 86 282 45 59 68 73 76 217

261 272 267 800 Expected Values

15 29 301 30 44 282 45 59 217

261 272 267 800

o

–

2 ( ) ( . )

.

( . )

.

( . )

.

( . )

.

( . )

. ( . )

.

( . )

.

( . )

.

( . )

. 1.927

2

p 0.7492