Question

QUESTION 1 In constructing a 95% confidence level estimate of the mean when the population standard deviation () is known what will be your score used in the formula? QUESTION 2 In constructing a 99% confidence level estimate of the mean when the population standard deviation (a) is known what will be your score used in the formula? HINT. Be sure to review page 236 "Finding Z scores from Known Areas - Special Cases and Tabel A-2. QUESTION 3 In constructing a 90% confidence level estimate of the mean when the population standard deviation (a) is known what will be your z score used in the formula? HINT: Be sure to review page 236 "Finding Z scores from Known Areas - Special Cases and Tabel A-2. QUESTION 4 In constructing a 98% confidence level estimate of the mean when the population standard deviation () is known what will be your z score used in the formula? QUESTION 5 In constructing a 95% confidence level estimate of the mean when the population standard deviation (a) is unknown and the sample size is 40, what will be your t score used in the formula? QUESTION 6

Answer 1

QUESTION-1:

SOLUTION:

When the population standard deviation (σ) is known,

a 95% confidence level estimate of the mean

=* 20.025,1 + =* 20.025) Vn

where $z_{0.025}$ is the upper 0.025 point of a standard normal distribution i.e.

P(Z > 20.025) = 0.025

1- P(Z < 20.025) = 0.025

= P(Z < 20.025) = 1-0.025 = 0.975

(20.025) = 0.975

(found from standard normal table)

= 20.025 = Φ-1 (0.975) = 1.95996

(ans)

= 0.0% = 1.95996

QUESTION-2:

SOLUTION:

A 99% confidence level has significance level of
_a=0.01
and critical value is,
_{Za/2 = 2.575}

Therefore, we should used z-score = 2.575

Note: if you want z-score rounded to two decimal places then it is equal to z = 2.58

QUESTION-3:

SOLUTION:

90% confidence interval

Here $\alpha$ = 10%

$\sigma$ Known, we can use normal distribution.

Z-score to be used = Z_0.05

Where Z_0.05 is defined such that. P(Z>Z_0.05) = 0.05.

From Z-table P(Z>1.645) = 0.05.

Hence, Z-score used = 1.645.

QUESTION-4:

SOLUTION:

In constructing a 98% confidence level estimate of the mean when the population standard deviation ($\sigma$) is known.

So, the Z-score we used is 2.33.

Because, here

a= 0,02

And we want to find the confidence interval to mean. So we need Z   _{$\alpha /2$} score.

That is Z_0.01 = 2.33 Answer.

QUESTION-5:

SOLUTION:

here degree of freedom =n-1=39

for 39 degree of freedom and 95% confidence level ; critical value of t score = -/+ 2.023

QUESTION-6:

SOLUTION:

df = 35 - 1 = 34

T score for 99% confidence interval = t_0.005,34 = 2.728

QUESTION-7:

SOLUTION:

df = 30 - 1 = 29

T score for 90% confidnece interval = t_0.05,29 = 1.699

QUESTION-8:

SOLUTION:

df = 33 - 1 = 32

T score for 98% confidnece interval = t_0.01,32 = 2.449

QUESTION-9:

SOLUTION:

At 95% confidence interval the critical value is t_{0.025, 39} = 2.023

The upper limit of the 95% confidence interval for population mean is

$\bar x$ + t_{0.025, 39} * s/$\sqrt n$

= 110.27 + 2.023 * 18.95/$\sqrt 40$

= 110.27 + 6.06

= 116.33

QUESTION-10:

SOLUTION:

$\widehat p$ = 35/200 = 0.175

At 95% confidence interval the critical value is z_0.025 = 1.96

The upper limit of the 95% confidence interval for population proportion is

$\widehat p$ + z_0.025 * sqrt($\widehat p$(1 - $\widehat p$ )/n)

= 0.175 + 1.96 * sqrt(0.175 * (1 - 0.175)/200)

= 0.175 + 0.053

= 0.228

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