Question

A web-based company has a goal of processing 95 percent of its orders on the same day they are received. If 485 out of the next 500 orders are processed on the same day, would this prove that they are exceeding their goal, using α = .025? Steps: 1. Open Minitab on your computer. 2. There is no data file for this example. The information above contains summarized data (take info from story and plug into Minitab) (a) H0:p ≤ .95 versus HA: p > .95. Choose the right option. a. Reject H0 if p-value < alpha b. Reject H0 if p-value > alpha a b

Answer 1

Answer:

The null hypothesis was rejected.

Conclusion: The proportions of orders completed on the day of receiving is more than 95%.

Step-by-step explanation:

The hypothesis can be defined as:

H₀: The proportions of orders completed on the day of receiving is not more than 95%, i.e. p ≤ 0.95.

Hₐ: The proportions of orders completed on the day of receiving is more than 95%, i.e. p > 0.95.

The significance level of the test is α = 0.025.

The sample size is, n = 500.

As the sample size is large, the sampling distribution of sample proportion can be approximated by the Normal distribution.

The mean and standard deviation of this distribution are:

$\mu=p\\\sigma=\sqrt{\frac{p(1-p)}{n}}$

The test statistic is:

$z=\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n}} }$

The sample proportion is:

$\hat p=\frac{X}{n}=\frac{485}{500}=0.97$

Compute the test statistic as follows:

$z=\frac{0.97-0.95}{\sqrt{\frac{0.95(1-0.95)}{500}} }=2.05$

The decision rule is:

If the p-value is less than the significance level α then the null hypothesis is rejected.

Compute the p-value as follows:

$p-value=P(Z>2.05)\\=1-P(Z$

*Use a z-table.

The p-value = 0.0202 < α = 0.025.

The null hypothesis will be rejected.

Conclusion:

As the null hypothesis was rejected at 2.5% level of significance, it can be concluded that the proportions of orders completed on the day of receiving is more than 95%.

Answer 2

Answer:

Option a. Reject H0 if p-value < alpha

Step-by-step explanation:

We are given that a web-based company has a goal of processing 95 percent of its orders on the same day they are received. And 485 out of the next 500 orders are processed on the same day.

We are given that $\alpha$ = 0.025 .

Also, Null hypothesis, $H_0$ : p <= 0.95

Alternate Hypothesis, $H_1$ : p > 0.95

Our decision rule is given by;

If P-value is less than the significance level ($\alpha$) ⇒ reject null hypothesis $(H_0)$

If P-value is more than the significance level ($\alpha$) ⇒ accept null hypothesis $(H_0)$ .

So, option (a) is correct.