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Nimfa-mama [501]
3 years ago
10

According to a survey by the Better Sleep Council, 33% of people admit to dozing off at their workplace. Assume this proportion

represents the true proportion all workers who doze off at their workplace. If a random sample of 100 workers is selected from this population, determine the standard error of the proportion.
Mathematics
1 answer:
Lorico [155]3 years ago
6 0

Answer:

Standard error of the proportion   = 0.04702

Step-by-step explanation:

<u>Explanation</u>:-

According to a survey by the Better Sleep Council, 33% of people admit to dozing off at their workplace

The sample proportion p = 33% = 0.33

q = 1-p = 1-0.33 = 0.67

Given a random sample of 100 workers is selected from this population

Given sample size 'n' = 100

The standard error of the Proportion  is determined by

                                                                                     S.E = \frac{\sqrt{pq} }{\sqrt{n} }

                                                                                      S.E = \frac{\sqrt{0.33 X0.67} }{\sqrt{100} }

<u>Final answer</u>:-

  Standard error of the proportion   = 0.04702

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(1 point) The effectiveness of a new bug repellent is tested on 18 subjects for a 10 hour period. Based on the number and locati
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Answer:    

z=1.59

If we compare the p value and the significance level given \alpha=0.025 we see that p_v>\alpha so we can conclude that we fail to reject the null hypothesis, so we can conclude that the mean repellency of the new bug repellent is greater than 89% at 0.025 of signficance.    

Step-by-step explanation:    

1) Data given and notation    

\bar X=0.92 represent the mean effectiveness of a new bug repellent for the sample    

\sigma=0.08 represent the population standard deviation for the sample    

n=18 sample size    

\mu_o =0.89 represent the value that we want to test  

\alpha=0.025 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)    

p_v represent the p value for the test (variable of interest)

State the null and alternative hypotheses.    

We need to conduct a hypothesis in order to determine if the mean repellency of the new bug repellent is greater than 89% or 0.89, the system of hypothesis would be:    

Null hypothesis:\mu \leq 0.89    

Alternative hypothesis:\mu > 0.89    

We don't know the population deviation, and the sample size <30, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:    

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)    

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic    

We can replace in formula (1) the info given like this:    

t=\frac{0.92-0.89}{\frac{0.08}{\sqrt{18}}}=1.59    

Calculate the P-value    

First we need to calculate the degrees of freedom given by:

df=n-1=18-1=17

Since is a one-side upper test the p value would be:    

p_v =P(t_{17}>1.59)=0.065

In Excel we can use the following formula to find the p value "=1-T.DIST(1.59;17;TRUE)"  

Conclusion    

If we compare the p value and the significance level given \alpha=0.025 we see that p_v>\alpha so we can conclude that we fail to reject the null hypothesis, so we can conclude that the mean repellency of the new bug repellent is greater than 89% at 0.025 of signficance.    

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