Question

7.1.37 Question Help A security expert claims that more than 15% of all homeowners have a home security alarm. State Ho and H, in words and in symbols. Then determine whether the hypothesis test for this claim is left-tailed, right-tailed, or two-tailed. Explain your reasoning State the null hypothesis in words and in symbols. Choose the correct answer below. O A. The null hypothesis expressed in words is, "the proportion of all homeowners who own a home security alarm is at least 0.15." The null hypothesis is expressed symbolically as, "Ho: p20.15." OB. The null hypothesis expressed in words is, "the proportion of all homeowners who own a home security alarm is not 0.15." The null hypothesis is expressed symbolically as, "Ho: p=0.15." O C. The null hypothesis expressed in words is, "the proportion of all homeowners who own a home security alarm is at most 0.15." The null hypothesis is expressed symbolically as, "Hop 50.15." OD. The null hypothesis expressed in words is, "the proportion of all homeowners who own a home security alarm is 0.15." The null hypothesis is expressed symbolically as, "Ho: p=0.15."

Answer 1

1) Answer C. The null hypothesis expressed in words is, "the proportion of all homeowners who own a home security is at most 0.15". The null hypothesis is expressed symbolically as "Ho : p ≤ 0.15."

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2) Answer B. The null hypothesis expressed in words is, "the mean length of a baseball team's game is at most 1.2 hours". The null hypothesis is expressed symbolically as "Ho : µ ≤ 1.2."

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3) Right tailed critical value, z crit = ABS(NORM.S.INV(0.02)) = 2.05

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4) x̅ = 48.6, σ = 3.43, n = 58

Null and Alternative hypothesis: Answer C.

Ho : µ = 50

H1 : µ ≠ 50

Critical value :

Two tailed critical value, z crit = ABS(NORM.S.INV(0.05/2)) = +/- 1.960

Reject Ho if z < -1.96 or if z > 1.96

Test statistic:

z = (x̅- µ)/(σ/√n) = (48.6 - 50)/(3.43/√58) = -3.1085

p-value :

p-value = 2*(1-NORM.S.DIST(ABS(-3.1085, 1) = 0.0019

Decision:

p-value < α, Reject the null hypothesis

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5) x̅ = 294, σ = 40, n = 79

Null and Alternative hypothesis: Answer C.

Ho : µ ≤ 285

H1 : µ > 285 (claim)

Critical value :

Right tailed critical value, z crit = ABS(NORM.S.INV(0.06) = 1.555

Reject Ho if z > 1.555

Test statistic:

z = (x̅- µ)/(σ/√n) = (294 - 285)/(40/√79) = 1.9998

p-value :

p-value = 1- NORM.S.DIST(1.9998, 1) = 0.0228

Decision:

p-value < α, Reject the null hypothesis

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6) x̅ = 900, σ = 129, n = 60

Null and Alternative hypothesis: Answer F.

Ho : µ ≤ 880

H1 : µ > 880 (claim)

Critical value :

Right tailed critical value, z crit = ABS(NORM.S.INV(0.1) = 1.282

Reject Ho if z > 1.282

Test statistic:

z = (x̅- µ)/(σ/√n) = (900 - 880)/(129/√60) = 1.2009

p-value :

p-value = 1- NORM.S.DIST(1.2009, 1) = 0.1149

Decision:

p-value > α, Do not reject the null hypothesis