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r-ruslan [8.4K]
3 years ago
10

rachel owns a cupcake shop. last week she used 42.5 pounds of sugar making cupcakes. this week she used 26 pounds of sugar. find

the percent change in the amount of sugar used. classify as a percent increase or decrease
Mathematics
1 answer:
Paha777 [63]3 years ago
3 0

Answer:

decreased by 39%

Step-by-step explanation:

deacreased by

42.5-26= 16.5

16.5÷42.5=38.82352941

rounded to 39%

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Consider a drug testing company that provides a test for marijuana usage. Among 317 tested​ subjects, results from 25 subjects w
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Answer:

We conclude that we fail to reject H_0 as there is not sufficient evidence to warrant support of the claim that less than 10 percent of the test results are wrong.

Step-by-step explanation:

We are given that among 317 tested​ subjects, results from 25 subjects were wrong.

We have to test the claim that less than 10 percent of the test results are wrong.

<u><em>Let p = proportion of subjects that were wrong.</em></u>

So, Null Hypothesis, H_0 : p = 10%     {means that 10 percent of the test results are wrong}

Alternate Hypothesis, H_A : p < 10%     {means that less than 10 percent of the test results are wrong}

The test statistics that would be used here <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 = sample proportion of test results that were wrong = \frac{25}{317} = 0.08

           n = sample of tested subjects = 317

So, <u><em>test statistics</em></u>  =  \frac{0.08-0.10}{\sqrt{\frac{0.08 (1-0.08)}{317} } }

                              =  -1.31

The value of z test statistics is -1.31.

Now, the P-value of the test statistics is given by the following formula;

                P-value = P(Z < -1.31) = 1 - P(Z \leq 1.31)

                              = 1 - 0.9049 = <u>0.095</u>

<u><em>Now, at 0.05 significance level the z table gives critical value of -1.645 for left-tailed test.</em></u><em> Since our test statistics is more than the critical value of z as -1.31 > -1.645, 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 we fail to reject H_0 as there is not sufficient evidence to warrant support of the claim that less than 10 percent of the test results are wrong.

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