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Kryger [21]
3 years ago
6

A market research company wishes to know how many energy drinks adults drink each week. They want to construct a 80% confidence

interval for the mean and are assuming that the population standard deviation for the number of energy drinks consumed each week is 0.9. The study found that for a sample of 164 adults the mean number of energy drinks consumed per week is 7.9. Construct the desired confidence interval. Round your answers to one decimal place.
Mathematics
1 answer:
Diano4ka-milaya [45]3 years ago
7 0

Answer:

The confidence interval = (7.8 , 8.0)

Step-by-step explanation:

Confidence Interval formula =

Mean ± z × Standard deviation/√n

Mean = 7.9

Standard deviation = 0.9

n = number of samples = 164

z = z score of an 80% confidence interval = 1.282

Confidence Interval = 7.9 ± 1.282 × 0.9/√164

= 7.9 ± 0.0900966432

Confidence Interval

= 7.9 - 0.0900966432

= 7.8099033568

Approximately to 1 decimal place = 7.8

7.9 + 0.0900966432

= 7.9900966432

Approximately to 1 decimal place = 8.0

Therefore, the confidence interval = (7.8 , 8.0)

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Answer:

z=\frac{0.36 -0.39}{\sqrt{\frac{0.39(1-0.39)}{100}}}=-0.615  

The p value for this case would be:

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For this case since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true proportion is not different from 0.39

Step-by-step explanation:

Information given

n=100 represent the random sample taken

X=36 represent the number of people that take E supplement

\hat p=\frac{36}{100}=0.36 estimated proportion of people who take R supplement

p_o=0.39 is the value that we want to test

\alpha=0.05 represent the significance level

z would represent the statistic

p_v represent the p value

Hypothesis to test

We want to test if the true proportion is equatl to 0.39 or not, the system of hypothesis are.:  

Null hypothesis:p=0.39  

Alternative hypothesis:p \neq 0.39  

The statistic is given by:

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

Replacing the info we got:

z=\frac{0.36 -0.39}{\sqrt{\frac{0.39(1-0.39)}{100}}}=-0.615  

The p value for this case would be:

p_v =2*P(z  

For this case since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true proportion is not different from 0.39

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Answer:

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Step-by-step explanation:

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