A sample of 49 sales receipts from Apple store has the sample mean of $450 and p

Question

A sample of 49 sales receipts from Apple store has the sample mean of $450 and population standard deviation of $70. Use these values to test whether or not the mean of sales at the Apple store would be different from $430.

a) Find 95% Confidence Interval for a mean.

b)  Hypothesis Testing with significance level a = 0.05.

Step 1. State the null and alternative hypotheses.

Step 2. State Decision Rule for both p-value and critical value methods

Decision Rule for p-value method:

Decision Rule for Critical Value Method:

Step 3: Find Test Statistic:

Step 4: Conclusion

a. By p-value method

b. By critical value method

Explanation / Answer

Solution

Let X = sales receipts from Apple store.

We assume X ~ N(µ, ?2),where ? = population standard deviation of $70 [given]

Part (a)

100(1 - ?) % Confidence Interval for ?, when ? known is: Xbar ± (Z? /2)?/?n   where

Xbar = sample mean, Z? /2 = upper (? /2)% point of N(0, 1), ? = population standard deviation and n = sample size.

So, 95% [i.e., ? = 0.05]Confidence Interval for a mean = 450 ± (1.96 x 70)/?49

= 450 ± 19.6

= [430.4, 469.6] ANSWER 1

Part (b)

Hypotheses:

Null H0: µ = µ0 = 430   Vs Alternative HA: µ ? 430 ANSWER 2

Decision Rule

p-value method: Reject H0, if p-value < ? ANSWER 3

Critical Value Method: Reject H0, if | Zcal | > Zcrit. ANSWER 4

Test statistic:

Z = (?n)(Xbar - µ0)/?, where n = sample size; Xbar = sample average; ? = known population standard deviation.

= 7(450 - 430)/70

= 2 ANSWER 5

Critical Value and p-value given ? = 0.05 (5%)

Under H0, Z ~ N(0, 1)

Critical value = upper (?/2)% point of N(0, 1) = 1.96.

p-value = P(Z > |Zcal|) = P(Z > |2|) = 0.0456

Zcrit and p-value are found using Excel Function:

Decision and Conclusion::

Since | Zcal | > Zcrit, H0 is rejected.

Since p-value < ?. H0 is rejected.

There is sufficient evidence to support the claim that the mean of sales at the Apple store would be different from $430.

DONE

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