# 0.2 Practice tests (1-4) and final exams  (Page 22/36)

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8 . The 99% confidence interval, because it includes all but one percent of the distribution. The 95% confidence interval will be narrower, because it excludes five percent of the distribution.

## 8.2: confidence interval, single population mean, standard deviation unknown, student’s t

9 . The t -distribution will have more probability in its tails (“thicker tails”) and less probability near the mean of the distribution (“shorter in the center”).

10 . Both distributions are symmetrical and centered at zero.

11 . df = n – 1 = 20 – 1 = 19

12 . You can get the t -value from a probability table or a calculator. In this case, for a t -distribution with 19 degrees of freedom, and a 95% two-sided confidence interval, the value is 2.093, i.e.,
The calculator function is invT(0.975, 19).

13 .
98.4 ± 0.14 = (98.26, 98.54).
The calculator function Tinterval answer is (98.26, 98.54).

14 . ${t}_{\frac{\alpha }{2}}=2.861.$ The calculator function is invT(0.995, 19).
$EBM={t}_{\frac{\alpha }{2}}\left(\frac{s}{\sqrt{n}}\right)=\left(2.861\right)\left(\frac{0.3}{\sqrt{20}}\right)=0.192$
98.4 ± 0.19 = (98.21, 98.59). The calculator function Tinterval answer is (98.21, 98.59).

15 . df = n – 1 = 30 – 1 = 29.

98.4 ± 0.11 = (98.29, 98.51). The calculator function Tinterval answer is (98.29, 98.51).

## 8.3: confidence interval for a population proportion

16 . ${p}^{\prime }=\frac{280}{500}=0.56$
${q}^{\prime }=1-{p}^{\prime }=1-0.56=0.44$
$s=\sqrt{\frac{pq}{n}}=\sqrt{\frac{0.56\left(0.44\right)}{500}}=0.0222$

17 . Because you are using the normal approximation to the binomial, ${z}_{\frac{\alpha }{2}}=1.96$ .
Calculate the error bound for the population ( EBP ):

Calculate the 95% confidence interval:
0.56 ± 0.0435 = (0.5165, 0.6035).
The calculator function 1-PropZint answer is (0.5165, 0.6035).

18 . ${z}_{\frac{\alpha }{2}}=1.64$

0.56 ± 0.03 = (0.5236, 0.5964). The calculator function 1-PropZint answer is (0.5235, 0.5965)

19 . ${z}_{\frac{\alpha }{2}}=2.58$

0.56 ± 0.05 = (0.5127, 0.6173).
The calculator function 1-PropZint answer is (0.5028, 0.6172).

20 . EBP = 0.04 (because 4% = 0.04)
${z}_{\frac{\alpha }{2}}=1.96$ for a 95% confidence interval

You need 601 subjects (rounding upward from 600.25).

21 .
You need 577 subjects (rounding upward from 576.24).

22 .
You need 1,068 subjects (rounding upward from 1,067.11).

## 9.1: null and alternate hypotheses

23 . H 0 : p = 0.58
H a : p ≠ 0.58

24 . H 0 : p ≥ 0.58
H a : p <0.58

25 . H 0 : μ ≥ $268,000 H a : μ <$268,000

26 . H a : μ ≠ 107

27 . H a : p ≥ 0.25

## 9.2: outcomes and the type i and type ii errors

28 . a Type I error

29 . a Type II error

30 . Power = 1 – β = 1 – P (Type II error).

31 . The null hypothesis is that the patient does not have cancer. A Type I error would be detecting cancer when it is not present. A Type II error would be not detecting cancer when it is present. A Type II error is more serious, because failure to detect cancer could keep a patient from receiving appropriate treatment.

32 . The screening test has a ten percent probability of a Type I error, meaning that ten percent of the time, it will detect TB when it is not present.

if the death of of the snow is my yard is normally distributed with the m is equals to 2.5 and what is the probability that a randomly chosen location with have a no that between 2.25 and 2.76
hey
Shubham
🤔
Iqra
hello
Sakshi
hii
Rushikesh
helow
why Statistics so hard
Mohd
ho geya solve
Sakshi
it's not hard
Sakshi
it is hard 😭
Mohd
solution?
Abdul
hii
it's just need to be concentrate
Akinyemi
exactly..... concentration is very important
Iqra
rewrite the question
what is the true statement about random variable?
A consumer advocate agency wants to estimate the mean repair cost of a washing machine. the agency randomly selects 40 repair cost and find the mean to be $100.00.The standards deviation is$17.50. Construct a 90% confidence interval for the mean.
pls I need understand this statistics very will is giving me problem
Sixty-four third year high school students were given a standardized reading comprehension test. The mean and standard deviation obtained were 52.27 and 8.24, respectively. Is the mean significantly different from the population mean of 50? Use the 5% level of significance.
No
Ariel
how do I find the modal class
look for the highest occuring number in the class
Kusi
the probability of an event occuring is defined as?
The probability of an even occurring is expected event÷ event being cancelled or event occurring / event not occurring
Gokuna
what is simple bar chat
Simple Bar Chart is a Diagram which shows the data values in form of horizontal bars. It shows categories along y-axis and values along x-axis. The x-axis displays above the bars and y-axis displays on left of the bars with the bars extending to the right side according to their values.
statistics is percentage only
the first word is chance for that we use percentages
it is not at all that statistics is a percentage only
Shambhavi
I need more examples
how to calculate sample needed
mole of sample/mole ratio or Va Vb
Gokuna
how to I solve for arithmetic mean
Yeah. for you to say.
James
yes
niharu
how do I solve for arithmetic mean
niharu
add all the data and divide by the number of data sets. For example, if test scores were 70, 60, 70, 80 the total is 280 and the total data sets referred to as N is 4. Therfore the mean or arthritmatic average is 70. I hope this helps.
Jim
*Tan A - Tan B = sin(A-B)/CosA CosB ... *2sinQ/Cos 3Q = tan 3Q - tan Q
standard error of sample