The owner of a local nightclub has recently surveyed a random sample of n = 250

Question

The owner of a local nightclub has recently surveyed a random sample of n = 250 customers of the club. She would now like to determine whether or not the mean age of her customers is over 30, if so, she plans to alter the entertainment to appeal to an older crowd. If not, no entertainment changes will be made. Assume that the ages of her clientele are known to have a standard deviation of 5 years. If she wants to be 98% confident in her decision, what rejection region should she use? a. Reject H_0 if z > 2.33 b. Reject H_0 if z > 2.05 c. Reject H_0 if z

Explanation / Answer

given Sample size, n=250
Sample mean, x =30.45
Sample standard deviation ,=5
Let’s test whether or not the mean age of her customers is over 30
Null hypothesis: >30
Alternate hypothesis: 30

a) 98% ci critical value from standard z table its value is 2.33

Reject H0 : if Zo greater than 2.33

so option a is correct

b)
Test statistic
z=(x - )/(/n)
=(30.45- 30)/(5/250)
=(30.45- 30)/(5/250)
=0.14230249470758
=0.142 (Rounded to the three decimal places)
Since the level of significance is 0.01,
Critical value z stastic is P(z)=0.0778
so answer is option c

c) as alpha =0.02 the best conclusion is do not reject h0 as its alpha value is less greater than p value do not reject it

option a

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