Question

is found to be 10 A simple random sample of size nis drawn from a population that is normally distributed. The sample mean is found to be 112, and the sample standard deviation (a) Construct a 90% confidence interval about the sample sen, 29. (b) Construct a 96confidence interval about if the sample size is 11 (c) Construct a 70% condence interval about the sample sen, 29. (d) Could we have computed the confidence intervals in parts (He) if the population had not been normally distributed? Click the icon to view the table of reas under the distribution (a) Construct a 96% confidence interval about the sample size, n, 29 Lower bound: :Upper bound: Use ascending order. Hound to one decimal place as needed.) (b) Construct a confidence interval about the sample size is 11 Lower bound Upper bound: (Use ascending order. Round to one decimal place as needed) How does decreasing the sample size affect the margin of error ? OAAs the sample size decreases, the margin of error decreases B. As the sample size decreases, the margin of error stays the same OC. As the sample size decreases, the margin of error increases (c) Construct a 70% confidence interval about if the sample size, n, is 29 Lower bound Upper bound: (Use ascending order. Round to one decimal place as needed) Compare the results to those obtained in part (a). How does decreasing the level of confidence affect the size of the margin of error, E? O A As the percent confidence decreases the size of the interval decreases OB. As the percent confidence decreases the size of the interval stays the same Click to select your answers MacBook Pro

Answer 1

given data are,

sample mean ($\bar{x}$) = 112

sample sd (s) = 10

a).sample size (n) = 29

degrees of freedom = (n-1) = (29-1) = 28

t critical value for df = 28, 95 % confidence level , both tailed test be:-

$t^*=2.048$ [ from t distribution table ]

margin of error(E):-

$= [t^**\frac{s}{\sqrt{n}}]$

$=[2.048*\frac{10}{\sqrt{29}}]=3.8$

the 95 % confidence interval for population mean be:-

$\bar{x}\pm E$

$=112\pm 3.8$

$=(108.2,115.8)$

lower bound	108.2
upper bound	115.8

b).sample size (n) = 11

degrees of freedom = (n-1) = (11-1) = 10

t critical value for df = 10,95 % confidence level , both tailed test be:-

$t^*=2.228$ [ from t distribution table ]

margin of error(E):-

$= [t^**\frac{s}{\sqrt{n}}]$

$=[2.228*\frac{10}{\sqrt{11}}]=6.7$

the 95 % confidence interval for population mean be:-

$\bar{x}\pm E$

$=112\pm 6.7$

$=(105.3,118.7)$

lower bound	105.3
upper bound	118.7

INTERPRETATION:-

as the sample size decreases, the margin of error increases.

c).sample size (n) = 29

degrees of freedom = (n-1) = (29-1) = 28

t critical value for df = 28, 70 % confidence level , both tailed test be:-

$t^*=1.056$ [ from t distribution table ]

margin of error(E):-

$= [t^**\frac{s}{\sqrt{n}}]$

$=[1.056*\frac{10}{\sqrt{29}}]=2.0$

the 70 % confidence interval for population mean be:-

$\bar{x}\pm E$

$=112\pm 2$

$=(110,114)$

lower bound	110
upper bound	114

INTERPRETATION:-

as the percent confidence level decreases , the size of the interval decreases.

d). no, the population needs to be normally distributed.

[ because, here we are using t distribution..this t curve is a symmetric curve.. if the population is not normal ..then we will not be able to use this distribution ]

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