Researchers at The National Gallery in Ottawa have been collecting preference data on their permanent collection for over a decade. This data shows the average liking rating for art in the permanent collection to be 41 with a standard deviation of 3.1. Their visitor demographics seems to be changing and the research team wants to know if this year's visitors (n = 387) who have an average liking rating of 45 is different from the liking rating of past visitors.

 

Answer:

There is sufficient statistical evidence, with 95% confidence level, for affirm that:

The liking rating for this year is different from the liking rating of past visitors.

Step-by-step explanation

?TEST FOR THE MEAN OF A NORMAL DISTRIBUTION, VARIANCE KNOWN

μ=41  

σ=3.1  

n=387

x=45  

?(Population mean)(Population standard deviation)(Sample size)(Sample mean)??

α=0.05?(Significance level)??

Choice of statistic distribution

Ifx1?,x2?,...,xn?is a random sample of sizentaken from a population (either finite or infinite) with meanμand finite varianceσ2,and ifxis the sample mean, the limiting form of the distribution ofz=σ/n?x−μ?asn→∞,is thestandard normal distribution

??Hypothesis approach

Null hypothesis:   H0?:μ=41

Alternative hypothesis:?  H1?:μ?=41  ?(Also calledHa?)??Since the alternative hypothesis contains the "not equal to" symbol, this istwo-tailed test.??Significance level

α=0.05??

Critical value of test statistic,zα/2

?α=0.05

α/2?=20.05?

a/2=0.0250

zα/2?

value is thez−value having an area ofα/2 (0.0250)to the right. The cumulative area to the left is1−α/2=1−0.0250=0.9750

?Calculus ofzα/2 using the cumulative standard normal distribution table

We search through the probabilities to find the value that corresponds to0.9750.

..z....1.71.81.92.02.1..

.?0.00...0.95540.96410.97130.97720.9821..

.?0.01...0.95640.96490.97190.97780.9826...

?0.02...0.95730.96560.97260.97830.9830...

?0.03...0.95820.96640.97320.97880.9834...

?0.04...0.95910.96710.97380.97930.9838...

?0.05...0.95990.96780.97440.97980.9842...?

0.06...0.96080.96860.97500.98030.9846...

?0.07...0.96160.96930.97560.98080.9850...

?0.08...0.96250.96990.97610.98120.9854..

?0.09...0.96330.97060.97670.98170.9857.

?We find0.9750exactly. Therefore:

zα/2?=1.9+0.06

zα/2?=1.96???

For a two-tailed test, the critical values are:−zcritical?=−1.96andzcritical?=1.96.?Decision rule using the test statistic and critical value

Since H1?: μ?=41, the decision rule is:

If ztest?<−1.96orztest?>1.96:    Reject H0

If−1.96<ztest?<1.96:?        ?Fail to reject H0

???Test statistic

ztest?=σ/n?x−μ?

ztest?=3.1/387?45−41

?ztest?=3.1/19.672315574

?ztest?=0.15764

?ztest?=25.38?

Decision using the test statistic and critical value

We compare the value ofztest with the critical valuezcritical?.

ztest?(25.38)>zcritical?(1.96)

Decision:RejectH0?.??

Conclusion:

There is sufficient statistical evidence, with 95% confidence level, for affirm that:

The liking rating for this year is different from the liking rating of past visitors.