An insurance company collected data to estimate the mean value of personal prope

Question

An insurance company collected data to estimate the mean value of personal property held by apartment renters in Tuscaloosa. In a random sample of 45 renters, the average value of personal property was $14,280 and the standard deviation was $6,540.

1)  Three years ago the true mean value ofpersonal property owned by renters in Tuscaloosa was $13,050. Assuming that the true mean is unchanged from three years ago, find the probability that a future sample of 45 renters would result in a mean that is more extreme than the sample the insurance company just took. P(X > 14,280 | true mean = 13,050) What distribution must be used?

2) Based on this probability, would you conclude that the true mean has changed from three years ago? Or equivalently, is there sufficient evidence to conclude that the true mean has changed? Or equivalently, was the sample mean of $14,280 too close to $13,050 to call it unusual? Or was this value a rare event?

Explanation / Answer

1) As the sample size is large (n > 30), so we will use the normal distribution.

P(X > 14280)

= P((X - mean)/(sd/sqrt(n)) > (14280 - mean)/(sd/sqrt(n)))

= P(Z > (14280 - 13050)/(6540/sqrt(45)))

= P(Z > 1.26)

= 1 - P(Z < 1.26)

= 1 - 0.8962

= 0.1038

2) As the probability is not less than 0.05, so it is not unusual and there is not sufficient evidence to conclude that thr true mean has changed.

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