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Nadusha1986 [10]
3 years ago
13

45x50x39/87/2x633x91/20

Mathematics
1 answer:
UkoKoshka [18]3 years ago
8 0
I hope those x's mean multiplication, if so, here is your answer: 1452489.44
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Suppose that in a random selection of 100 colored candies, 21% of them are blue. The candy company claims that the percentage of
zhannawk [14.2K]

Answer:

The pvalue of the test is 0.0784 > 0.01, which means that at this significance level, we do not reject the null hypothesis that the percentage of blue candies is equal to 29%.

Step-by-step explanation:

The candy company claims that the percentage of blue candies is equal to 29%.

This means that the null hypothesis is:

H_{0}: p = 0.29

We want to test the hypothesis that this is true, so the alternate hypothesis is:

H_{a}: p \neq 0.29

The test statistic is:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

In which X is the sample mean, \mu is the value tested at the null hypothesis, \sigma is the standard deviation and n is the size of the sample.

0.29 is tested at the null hypothesis:

This means that \mu = 0.29, \sigma = \sqrt{0.29*0.71}

Suppose that in a random selection of 100 colored candies, 21% of them are blue.

This means that n = 100, X = 0.21

Value of the test statistic:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

z = \frac{0.21 - 0.29}{\frac{\sqrt{0.29*0.71}}{\sqrt{100}}}

z = -1.76

Pvalue of the test:

The pvalue of the test is the probability of the sample proportion differing at least 0.21 - 0.29 = 0.08 from the population proportion, which is 2 multiplied by the pvalue of Z = -1.76.

Looking at the z-table, z = -1.76 has a pvalue of 0.0392

2*0.0392 = 0.0784

The pvalue of the test is 0.0784 > 0.01, which means that at this significance level, we do not reject the null hypothesis that the percentage of blue candies is equal to 29%.

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Step-by-step explanation:

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(0,1)

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