In: Statistics and Probability
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
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