Question

The number of entrees purchased in a single order at a Noodles & Company restaurant has had an historical average of 2.05 entrees per order. On a particular Saturday afternoon, a random sample of 50 Noodles orders had a mean number of entrees equal to 2.4 with a standard deviation equal to 0.82. At the 5 percent level of significance, does this sample show that the average number of entrees per order was greater than expected?

(a) Choose the correct null and alternative hypotheses.
a. H0: μ ≥ 2.05 vs. H1: μ < 2.05
b. H0: μ ≤ 2.05 vs. H1: μ > 2.05
c. H0: μ = 2.05 vs. H1: μ ≠ 2.05

Answer 1

Answer: b. $H_0:\mu=2.05$ vs $H_1: \mu>2.05$

Step-by-step explanation:

Let $\mu$ be the population mean.

Given : The number of entrees purchased in a single order at a Noodles & Company restaurant has had an historical average of 2.05 entrees per order.

$\mu=2.05$

Claim : The average number of entrees per order was greater than expected.

i.e.$\mu>2.05$

Hence, the  correct null and alternative hypotheses for the given description:-

$H_0:\mu=2.05$

$H_1: \mu>2.05$   [Alternative hypothesis shows significance difference.]

Since , the alternative hypothesis is right-tailed , so the test is a right tailed test.

Test statistic : $z=\dfrac{2.4-2.05}{\dfrac{0.82}{\sqrt{50}}}\approx3.02$

p-value : P(z>3.02)=0.0012639

Since p-value (0.0012639) is less than the significance level (0.05) , so we reject the null hypothesis.

Conclusion : The average number of entrees per order was greater than expected