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love history [14]
1 year ago
10

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
Mathematics
1 answer:
Kamila [148]1 year ago
8 0

Answer:

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 = <u><em>proportion of orders processing on the same day they are received.</em></u>

SO, Null Hypothesis, H_0 : p \leq 0.90     {means that they are not exceeding their goal of 90%}

Alternate Hypothesis, H_0 : p > 0.90      {means that they are exceeding their goal of 90%}

The test statistics that would be used here <u>One-sample z test for</u> <u>proportions</u>;

                            T.S. =  \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }  ~ N(0,1)

where, \hat p = sample proportion of orders that are processed on the same day = \frac{434}{471} = 0.92

           n = sample of orders = 471

So, <u><em>the test statistics</em></u>  =  \frac{0.92-0.90}{\sqrt{\frac{0.92(1-0.92)}{471} } }

                                     =  1.599

The value of z test statistics is 1.599.

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

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 <u><em>we fail to reject our null hypothesis</em></u>.

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

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a

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     The number of defective is  k =  18

The null hypothesis is  H_o  :  p  =  0.08

The  alternative hypothesis is  H_a  :  p > 0.08

Generally the sample proportion is mathematically evaluated as

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Next we obtain the critical value of  \frac{ \alpha }{2} from the normal distribution table, the value is  

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Generally the standard of error is mathematically represented as

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         SE   =  \sqrt{\frac{0.09  (1 -  0.09)}{200} }

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The  margin of error is  

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The  95% confidence interval is mathematically represented as

     \r p  -  E  <  \mu <  p <  \r p  + E

=>   0.09 - 0.0397  <  \mu <  p < 0.09 + 0.0397

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