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Vaselesa [24]
3 years ago
7

Determine whether the statement describes a descriptive or inferential statistic. A survey of 3459 people revealed that 52% work

a full-time job; therefore it can be assumed that 52% of the U.S. population works a full-time job.
Mathematics
1 answer:
andriy [413]3 years ago
7 0

Answer:

Inferential statistics.

Step-by-step explanation:

We have been given a statement. We are asked to determine whether the given statement describes a descriptive or inferential statistic.

Statement: A survey of 3459 people revealed that 52% work a full-time job; therefore it can be assumed that 52% of the U.S. population works a full-time job.

We know that descriptive statistics represents the description about population using the data through numerical calculations, graphs or tables. While inferential statistics makes predictions about a population based on a sample of data taken from the population.

Since our statement gives the prediction about people, who work full-time in U.S. based on a survey of 3459 U.S. citizens, therefore, the given statement is inferential statistic.

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<u>Answer:</u>

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A sample of 100 workers located in Atlanta has an average daily work time of 6.5 hours with a standard deviation of 0.5 hours. A
Readme [11.4K]

Answer:

t=\frac{6.5-6.7}{\sqrt{\frac{0.5^2}{100}+\frac{0.7^2}{110}}}}=-2.398  

df = n_1 +n_2 -2= 100+110-2= 208

Since is a bilateral test the p value would be:

p_v =2*P(t_{208}

Comparing the p value with the significance level given \alpha=0.05 we see that p_v so we can conclude that we can reject the null hypothesis, and we have significant differences between the two groups at 5% of significance.

Step-by-step explanation:

Data given and notation

\bar X_{1}=6.5 represent the sample mean for Atlanta

\bar X_{2}=6.7 represent the sample mean for Chicago

s_{1}=0.5 represent the sample deviation for Atlanta

s_{2}=0.7 represent the sample standard deviation for Chicago

n_{1}=100 sample size for the group Atlanta

n_{2}=110 sample size for the group Chicago

t would represent the statistic (variable of interest)

\alpha=0.01 significance level provided

Develop the null and alternative hypotheses for this study?

We need to conduct a hypothesis in order to check if the meanfor atlanta is different from the mean of Chicago, the system of hypothesis would be:

Null hypothesis:\mu_{1}=\mu_{2}

Alternative hypothesis:\mu_{1} \neq \mu_{2}

Since we don't know the population deviations for each group, for this case is better apply a t test to compare means, and the statistic is given by:

t=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}} (1)

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Calculate the value of the test statistic for this hypothesis testing.

Since we have all the values we can replace in formula (1) like this:

t=\frac{6.5-6.7}{\sqrt{\frac{0.5^2}{100}+\frac{0.7^2}{110}}}}=-2.398  

What is the p-value for this hypothesis test?

The degrees of freedom are given by:

df = n_1 +n_2 -2= 100+110-2= 208

Since is a bilateral test the p value would be:

p_v =2*P(t_{208}

Based on the p-value, what is your conclusion?

Comparing the p value with the significance level given \alpha=0.05 we see that p_v so we can conclude that we can reject the null hypothesis, and we have significant differences between the two groups at 5% of significance.

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