Question

I am shopping for a dresser on KSL. I look at 30 dressers. I know the standard deviation of dresser prices

is $20, and the mean cost of the dressers I look at is $80. I had been told that dressers should be $50.

(a) Write down the null and alternative hypotheses.

(b) What is the value of the test statistic?

(c) What is the p-value? Interpret this value.

(d) What is the rejection region if = :04?

(e) Interpret the result - do we accept or reject the null hypothesis.

(f) Create a 95% condence interval for the true average cost of a dresser and interpret the condence

interval.

Answer 1

a)

Null hypothesis : Cost of dressers is $50 i.e. u = 50

Alternative hypothesis : Cost of dressers is not $50 i.e. u /= 50

b)

Since in this case the sample size is 30 which can be assumed to be large enough for normal approximation thorugh Central Limit Theorem.

Hence we can say that sample mean of cost of dressers () follow normal distribution with mean $80 and standard deviation $20.

Also the population standard deviation is known hence we can use the Z-test, so test statistic is given as :

= -8.215838363

c)

p value represents the probability of obtaining a value of the tested variable as extreme as the test statistic assuming that the null hypothesis is correct. A smaller value of p value implies that our assumption of null hypothesis being true is wrong and vice versa is the case when the p value of high. P value is compared against the level of significance to determine which hypothesis to accept. If p value is higher than alpha then we accept null hypothesis and reject it in case p value is lower than alpha

d)

If = 0.04

Then rejected region is determined by the following values in case of two tailed test as above.

&   

where Z corresponds to the table values of standard normal distribution.

In this case the above values equilate to :

- 2.0537 & 2.0537

Hence rejection region :

e)

The test statistic value (8.215838363) belongs to the rejection region

Hence we have enough evidence to state that the null hypothesis is rejected at 4% level of significance. Therefore the cost of dressers is not equal to $50

f)

True average value ( u ) = $80

Standard deviation ( s ) = $20

Sample size (n) = 30

The 95% confidence interval is given as :

( 72.84309 , 87.15691)

Interpretation :

A 95% confidence interval means that out of a set of average cost of dressers from a sample of size 30, 95% of those values are likely to fall within the given confidence interval.