A web-based company has a goal of processing 90 percent of its orders on the same day they are received. If 434 out of the next 471 orders are processed on the same day, would this prove that they are exceeding their goal, using α = .025? (a) H0: π ≤ .90 versus H1: π > .90. Choose the right option. Reject H0 if zcalc > 1.96 Reject H0 if zcalc < 1.96 a b (b) Calculate the test statistic. (Round your answer to 3 decimal places.) Test statistic 1.013 (c-1) The null hypothesis should be rejected. No Yes (c-2) The true proportion is greater than .90. No evidence to support Yes (c-3) The company is exceeding its goal. No evidence to support Yes

We conclude that a web-based company are not exceeding their goal of 90%.

Step-by-step explanation:

We are given that a web-based company has a goal of processing 90 percent of its orders on the same day they are received.

434 out of the next 471 orders are processed on the same day.

Let p = proportion of orders processing on the same day they are received.

SO, Null Hypothesis, : p 0.90     {means that they are not exceeding their goal of 90%}

Alternate Hypothesis, : p > 0.90      {means that they are exceeding their goal of 90%}

The test statistics that would be used here One-sample z test for proportions;

                            T.S. =    ~ N(0,1)

where, = sample proportion of orders that are processed on the same day = = 0.92

           n = sample of orders = 471

So, the test statistics  =  

                                     =  1.599

The value of z test statistics is 1.599.

Now, at 0.025 significance level the z table gives critical value of 1.96 for right-tailed test.

Since our test statistic is less than the critical value of z as 1.599 < 1.96, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that a web-based company are not exceeding their goal of 90%.

seems like d

step-by-step explanation:

d. 24+-0.114

step-by-step explanation:

the question requires us to calculate the approximate 95% confidence interval for the mean number of ounces of ketchup per bottle in the sample given the following statistics;

sample mean = 24

standard deviation, σ = 0.4

sample size , n = 49

the confidence interval for a population mean is calculated using the formula;

sample mean ± z-score*(σ/sqrt(n))

the z-score associated with a 95% confidence interval, from the standard normal tables, is;

1.96

substituting these values into the formula;

24 ± 1.96*(0.4/7)

= 24 ± 0.112

from the alternatives given, choice d is closest to the above expression;