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Ivanshal [37]
3 years ago
5

To better understand the financial burden students are faced with each term, the statistics department would like to know how mu

ch their ST201 students are spending on school materials on average. Let’s use our class data to calculate a 95% confidence interval to estimate the average amount ST201 students spend on materials each term. The average from our student survey is $248 and the number of students sampled is 90. Assume . State the question of interest. On average, how much do ST201 students spend on school materials each term? a. (1 point) Identify the parameter. b. Check the conditions. a. (2 points) Does the data come from a random sample? What are some potential biases about the way the data was collected? (1 point) Is the sample size large enough for distribution of the sample mean to be normal according to the rules for Central Limit Theorem?
Mathematics
1 answer:
Semmy [17]3 years ago
3 0

Answer:

Answer:

a).

The amount spent on school materials for each term of all ST201students

b).

a).

It is not a random sample. This looks like a convenience sampling and there is sampling bias. This sample is not representative of the entire population. Since it is not a random sample it is not appropriate to generalize the results to all students.

b).

The sample size is 80 which is greater than 30. It is large enough to assume normal distribution according to central limit theorem.

c).

mean: $617

z critical value at 95%: 1.96

standard error = σ/sqrt(n) =500/sqrt(80) = 55.9017

lower limit= mean-1.96*se = 617-1.96*55.9017=507.43

upper limit= mean+1.96*se = 617+1.96*55.9017=726.57

d).

The amount spent on school materials for each term for the 80 ST201students is $617. We are 95% confident that amount spent on school materials for each term of all ST201students falls in the interval ($507.43, $726.57).

Step-by-step explanation:

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Answer:

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p_v =P(t_{(49)}  

If we compare the p value and the significance level assumed \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the height of men actually its significant lower than 26 so then the specification is satisfied.

Step-by-step explanation:

Data given and notation  

\bar X=25.02 represent the sample mean

s=4.83 represent the sample standard deviation

n=50 sample size  

\mu_o =26 represent the value that we want to test

\alpha 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 check if the mean is at least 26 mpg, the system of hypothesis would be:  

Null hypothesis:\mu \geq 26  

Alternative hypothesis:\mu < 26  

If we analyze the size for the sample is > 30 but we don't know the population deviation so 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{25.02-26}{\frac{4.83}{\sqrt{50}}}=-1.435    

P-value

The first step is calculate the degrees of freedom, on this case:  

df=n-1=50-1=49  

Since is a one sided test the p value would be:  

p_v =P(t_{(49)}  

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