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3241004551 [841]
2 years ago
12

Direct Mailing Company sells computers and computer parts by mail. The company claims that more than 90% of all orders are maile

d within 72 hours after they are received. The quality control department at the company often takes samples to check if this claim is valid. A recently taken sample of 175 orders showed that 161 of them were mailed within 72 hours. Do you think the company’s claim is true? Use the 1% significance level. What type of problem is this?
Mathematics
1 answer:
lys-0071 [83]2 years ago
6 0

Answer:

We conclude that less than or equal to 90% of all orders are mailed within 72 hours after they are received which means the company's claim is not true.

Step-by-step explanation:

We are given that the company claims that more than 90% of all orders are mailed within 72 hours after they are received. The quality control department at the company often takes samples to check if this claim is valid.

A recently taken sample of 175 orders showed that 161 of them were mailed within 72 hours.

<u><em>Let p = percentage of all orders that are mailed within 72 hours.</em></u>

SO, Null Hypothesis, H_0 : p \leq 90%   {means that less than or equal to 90% of all orders are mailed within 72 hours after they are received}

Alternate Hypothesis, H_A : p > 90%   {means that more than 90% of all orders are mailed within 72 hours after they are received}

The test statistics that will be used here is <u>One-sample z proportion</u> <u>statistics</u>;

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

where, \hat p = proportion of orders that were mailed within 72 hours in a sample of 175 =  \frac{161}{175}\times 100  = 92%

           n = sample of orders = 175

So, <u><em>test statistics</em></u>  =  \frac{0.92-0.90}{{\sqrt{\frac{0.92(1-0.92)}{175} } } } }

                               =  0.975

The value of the test statistics is 0.975.

<em>Now at 1% significance level, the z table gives critical value of 2.3263 for right-tailed test. Since our test statistics is less than the critical value of z as 0.975 < 2.3263, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which </em><em><u>we fail to reject our null hypothesis</u></em><em>.</em>

Therefore, we conclude that less than or equal to 90% of all orders are mailed within 72 hours after they are received which means the company's claim is not true.

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In order to find the variance we need to find first the second moment given by:

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And replacing we got:

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Step-by-step explanation:

Previous concepts

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).  

Solution to the problem

LEt X the random variable who represent the number of defective transistors. For this case we have the following probability distribution for X

X         0           1           2         3

P(X)    0.92     0.03    0.03     0.02

We can calculate the expected value with the following formula:

E(X) = \sum_{i=1}^n X_i P(X_i)

And replacing we got:

E(X) = 0*0.92 + 1*0.03 +2*0.03 +3*0.02 = 0.1500

In order to find the variance we need to find first the second moment given by:

E(X^2) = \sum_{i=1}^n X^2_i P(X_i)

And replacing we got:

E(X^2) = 0^2*0.92 + 1^2*0.03 +2^2*0.03 +3^2*0.02 = 0.3300

The variance is calculated with this formula:

Var(X) = E(X^2) -[E(X)]^2 = 0.33 -(0.15)^2 = 0.3075

And the standard deviation is just the square root of the variance and we got:

Sd(X) = \sqrt{0.3075}= 0.5545

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