Apoll of 2,014 randomly selected adults showed that 93% of them own cell phones. the technology display

10 answers
Question:

Apoll of 2,014 randomly selected adults showed that 93% of them own cell phones. the technology display below results from a test of the claim that 91%of adults own cell phones. use the normal distribution as an approximation to the binomial distribution, and assume a 0.01 significance level to complete parts test of p=0.91 vs pnot =0.91 sample xn sample p 95% ciz-value p-value118812 , 0140.933962 (0.919708, 0.948217) 3.760.000

Answers

91%

Step-by-step explanation:

0.069

Step-by-step explanation:

The given data set is

0.92, 0.87, 0.88, 0.82, 0.82, 0.87, 0.97, 0.86, 0.89, 0.84, 0.81, 0.88, 0.77, 0.86, 0.93, 0.84, 0.72, 0.82, 0.74, 0.83, 0.93, 0.75, 0.79, 0.91, 0.84, 0.91, 0.88, 0.88, 0.83, 0.78, 0.99, 0.81, 0.78, 0.75, 0.82, 0.76, 0.82, 0.87, 0.91, 0.77, 0.72, 0.94, 0.71, 0.73, 0.81, 0.81, 0.86, 0.93, 0.93, 0.82.

Formula for mean:

[tex]Mean=\frac{\sum x}{n}[/tex]

Sum of all terms = 41.98

Mean of the data set is

[tex]Mean=\frac{41.98}{50}[/tex]

[tex]Mean=0.8396[/tex]

Formula for standard deviation for population:

[tex]\sigma=\sqrt{\frac{\sum (x-\overline{x})^2}{n}}[/tex]

Formula for standard deviation for sample:

[tex]\sigma=\sqrt{\frac{\sum (x-\overline{x})^2}{n-1}}[/tex]

[tex]\sigma=\sqrt{\frac{0.231792}{50-1}}[/tex]

[tex]\sigma=\sqrt{0.004730449}[/tex]

[tex]\sigma=0.06877826[/tex]

[tex]\sigma\approx 0.069[/tex]

Therefore, the standard deviation of the data set is 0.069.

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Explanation:

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Explanation:

C

Step-by-step explanation:

The scores in a recent statistics test are given in the frequency distribution below.

[tex]\left|\begin{array}{c|cc}$Scores&$Frequency\\---&--\\0-60&4\\61-70&10\\71-80& 12\\81-90&4\\91-10&5\\----&--\\$Total&35\end{array}\right|[/tex]

The relative frequency is calculated in the table below.

[tex]\left|\begin{array}{c|c|c}$Scores&$Frequency&$Relative Frequency\\---&--&-----\\0-60&4&\dfrac{4}{35}\times 100=11.43\% \\\\61-70&10&\dfrac{10}{35}\times 100=28.57\%\\\\71-80& 12&\dfrac{12}{35}\times 100=34.29\%\\\\81-90&4&\dfrac{4}{35}\times 100=11.43\%\\\\91-10&5&\dfrac{5}{35}\times 100=14.29\%\\----&--&---\\$Total&35&100\end{array}\right|[/tex]

Therefore, the relative frequency table is that in Option C.

Correct option: (3).

Step-by-step explanation:

The hypothesis to determine whether the voter party loyalty has changed since 1992 is:

H₀: The voter party loyalty has not changed since 1992, i.e. p = 0.91.

Hₐ: The voter party loyalty has changed since 1992, i.e. p ≠ 0.91.

The decision rule is:

If the 90% confidence interval consists of the null hypothesis value, 0.91 then the null hypothesis will be accepted or else it will be rejected.

The 90% confidence interval for the proportion of self-proclaimed Republicans is:

(0.7564, 0.8389)

As the null hypothesis value, 0.91 is not included in the interval, it implies that the null hypothesis is rejected.

Conclusion:

As the null hypothesis was rejected at 10% level of significance it can be concluded that the voter party loyalty has changed since 1992.

Now consider the confidence interval (0.7564 , 0.8389).

The interval included values less than 0.91.

This implies that the proportion of self-proclaimed Republicans who voted in the 2012 election for their party candidate is smaller than 0.91.

Thus, the correct option is (3).

I will be attaching files to explain the procedure

Step-by-step explanation:

From the word document, we can see the data was first arranged, and then a frequency distribution table was created, and cummulative frequency calculated.

The class interval used was 2.5 and the cummulative frequency was 83 ( this is the total amount of observations in the data).

The histogram was improvised and I will attach a picture of it.


[tex]An article in Technometrics (Vol. 19, 1977, p. 425) presents the following data on the motor fuel oc[/tex]

For the first question, I think the best answer would be the fourth choice.

The 95% confidence interval for the slope of the least squares regression line would be: (0.38, 0.42).

I hope my answer has come to your help. God bless and have a nice day ahead!

Hello there.

Do the levels of fear in children change over time? If so, in what ways? Very little research has been done on the prevalence and persistence of fears in children. Researchers surveyed a group of third and fourth grade children and asked them to rate their level of fear about a variety of situations. After two years the children again completed the same survey. The researchers calculated the mean fear rating for each child in both years and were interested in the relation between these ratings. The least-squares regression line for the data, with 89 mean fear ratings as the explanatory variable and 91 mean fear ratings as the response variable, is given below. 

Dependent variable is 91 means 

R squared = 27.4% 

S = 0.2374 with 94 ? 2 = 92 degrees of freedom
Variable Coefficient s.e. of coefficient t-ratio P-value
Constant 0.877917 0.1184 7.42 ¡Ü 0.0001
89 means 0.397911 0.0676 5.89 ¡Ü 0.0001
Approximately what is a 95% confidence interval for the slope of the least squares regression line?

D. (0.33, 0.47) 

I just substituted the x on all of them and only one got the corrct y which is A.

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