cell phones discussion around 150 words 1- Savvy Essay Writers | savvyessaywriters.net

Context

In 2015 college students at a large state university completed a survey about their academic and personal life. Questions ranged from “How many credits are you registered for this semester?” to “Would you define yourself as a vegetarian?”

The researcher randomly selected 325 students from the university; 312 responded to the survey.

Prompt

The cell-phone datafile is available in the Data section below.

Once again, here is the research question for this lab.

Based on a recent study, roughly 80% of college students in the U.S. own a smartphone. Is the proportion of smartphone owners lower at this university?

Respond to each of the following in your initial post.

1. State your hypotheses in symbolic form and in words. (The following should be clear in your answer: the population of interest and the meaning of the proportion p in terms of the variable Cell.)
2. StatCrunch uses a normal model to estimate the P-value probability. Verify that normality conditions are met.

3. Give your P-value and interpret its meaning as a conditional probability.

4.State a conclusion that answers the research question. Use a significance level of 5%. (Your answer should include the P-value and reference the population and the variable Cell.)

5. Use StatCrunch to conduct the hypothesis test.
Copy and paste the results (the StatCrunch output window) into your initial post.

One sample proportion hypothesis test:

Outcomes in : Cell
Success : YES
p : Proportion of successes
H₀ : p = 0.8
H_A : p < 0.8

Hypothesis test results:

Variable	Count	Total	Sample Prop.	Std. Err.	Z-Stat	P-value
Cell	0	310	0	0.022718473	-35.213634	<0.0001

ANSWER(S): { Hint }

1. 
   
   
   H
   
   
   0
   
   :
   
   p
   
   
   =
   
   
   0.80
   ,
   
   
   H
   
   
   a
   
   :
   
   p
   
   
   <
   
   
   0.80
   , where p is the proportion of students at this university who own a smart phone.
2. Both normality conditions are met:
   (i) The sample is random.
   (ii)
   
   n
   
   
   =
   
   
   312
   , and
   
   
   P
   
   
   0
   
   
   
   =
   
   
   0.80
   and so
   
   312
   
   
   
   (
   
   
   0.8
   
   
   )
   
   
   
   =
   
   
   249.6
   
   
   ≥
   
   
   10
   and also
   
   312
   
   
   
   (
   
   
   1
   
   
   −
   
   
   0.80
   
   
   )
   
   
   
   =
   
   
   62.4
   
   
   ≥
   
   
   10
   .

   So we may assume that the distribution of sample proportions is approximately normal.

3. 
4. The P-value is 0.2389. The P-value is the probability that in random samples of 312 students, 78% or fewer will own a smart phone if in actuality 80% of the students at the university own a smart phone.
5. The P-value is 0.2393 “>
   
   >
   0.05, so we fail to reject the null hypothesis.

   Even though the sample proportion of 0.78 is less than the hypothesize population proportion of 0.80, the difference is not statistically significant. This sample does not provide strong enough evidence to conclude that the proportion of students at this university who own a cell phone is less than 0.80 (which is also the national proportion).

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