Question

Find the indicated critical z value.

Find zα/2 for α = 0.08.

1.96
2.65
1.41
1.75

Answer 1

do you have a TI83 or TI84 calculator?

Answer 2

i do not

Answer 3

ok let me find another calculator for you

Answer 4

okay thank you

Answer 5

ok go here

http://onlinestatbook.com/2/calculators/inverse_normal_dist.html

type in "0.04" in the "area" box without quotes
leave "Mean" and "SD" alone

tell me what you get

Answer 6

btw, I got 0.04 from dividing alpha = 0.08 in half

Answer 7

okay I got 1.751

Answer 8

me too

Answer 9

can you help me with this?

Assume that women's heights are normally distributed with a mean of 63.6 inches and a standard deviation of 2.5 inches. If 90 women are randomly selected, find the probability that they have a mean height between 62.9 inches and 64.0 inches.

0.1739
0.0424
0.7248
0.9318

Answer 10

I'm horrible with standard deviations

Answer 11

First we need the z-score for the rawscore xbar = 62.9

\[\LARGE z = \frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\]
\[\LARGE z = \frac{62.9-63.6}{\frac{2.5}{\sqrt{90}}}\]
\[\LARGE z = ??\]

Answer 12

I got -2.6563132345

Answer 13

I think I'm wrong :(

Answer 14

that's correct

Answer 15

what's the z-score for the rawscore of xbar = 64.0 ?

Answer 16

1.5178932769

Answer 17

correct

Answer 18

so P(62.9 < xbar < 64.0) = P(-2.66 < Z < 1.52)

Answer 19

now you'll use a table or a calculator to compute P(-2.66 < Z < 1.52)

Answer 20

do you know how to do that? or no?

Answer 21

no :(

Answer 22

do you have a table with you?

Answer 23

a stats table that provides area under the standard normal curve

Answer 24

no all I have is the z score table

Answer 25

ok I'll be right back in a few minutes and show you how to use a table to answer this question. Then we can also go over using a calculator as well

Answer 26

okay thank you so much. my teacher doesn't help me nearly as much.

Answer 27

ok I'm back

Answer 28

the z score table you have, is that accessible online? If so, can I get the link?

if not, then I'll find another table we can use

Answer 29

https://www.stat.tamu.edu/~lzhou/stat302/standardnormaltable.pdf

Answer 30

ok great

Answer 31

first let's find P(Z < -2.66)

Answer 32

locate the row that starts with -2.6

then find the column that has 0.06

the row and column combine to form -2.66

the value in that row/column combo is the approximate area under the curve to the left of z = -2.66

Answer 33

see attached to see what I mean

Answer 34

.00391 ?

Answer 35

correct

Answer 36

what is P(Z < 1.52) equal to?

Answer 37

.93574

Answer 38

good

Answer 39

subtract the two values to get the area between z = -2.66 and z = 1.52

Answer 40

[ P(Z < 1.52) ] - [ P(Z < -2.66) ]

[ 0.93574 ] - [ 0.00391 ]

0.93183

Answer 41

That means P(-2.66 < Z < 1.52) = 0.93183

and P(62.9 < xbar < 64.0) = 0.93183

Answer 42

it seems so much better now thanks

Answer 43

np

Answer 44

would this be a continuation of the first one?

Find the critical value zα/2 that corresponds to a 98% confidence level.

2.575
1.75
2.33
2.05

Answer 45

Confidence level = 98%

alpha = 1-(confidence level)

alpha = 1-0.98

alpha = 0.02

Answer 46

alpha = 0.02

alpha/2 = 0.02/2

alpha/2 = 0.01

Answer 47

you will use that calculator I gave you and type in 0.01 (where you typed in 0.04)

Answer 48

i got 2.23 so 2.33

Answer 49

2.32 I mean

Answer 50

2.327

Answer 51

which becomes 2.33, yep

Answer 52

This is like the other one we just did right?

For women aged 18–24, systolic blood pressures (in mm Hg) are normally distributed with a mean of 114.8 and a standard deviation of 13.1. If 23 women aged 18–24 are randomly selected, find the probability that their mean systolic blood pressure is between 119 and 122.


0.3343
0.9341
0.0833
0.0577

Answer 53

yes just with different numbers

Answer 54

rawscore: xbar = 119
z-score: z = ???

Answer 55

I did the whole thing and got it!

in the end I got 0.93407

Answer 56

But I don't know how to do this one with no mean or standard deviation provided
Can you help me?

Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p.

A survey of 865 voters in one state reveals that 408 favor approval of an issue before the legislature. Construct the 95% confidence interval for the true proportion of all voters in the state who favor approval.

0.438 < p < 0.505
0.471 < p < 0.472
0.435 < p < 0.508
0.444 < p < 0.500

Answer 57

Use the formulas:


\[\Large L = \hat{p} - z*\sqrt{\frac{\hat{p}*(1-\hat{p})}{n}}\]

\[\Large U = \hat{p} + z*\sqrt{\frac{\hat{p}*(1-\hat{p})}{n}}\]


where...

L = Lower endpoint of confidence interval
U = Upper endpoint of confidence interval
The confidence interval is of the form (L,U) which can be written as L < p < U

\(\Large \hat{p}\) is the sample proportion (the symbol is read has "p hat")
z = critical value based on the confidence level
n = sample size

Answer 58

okay thanks trying it now

Answer 59

Okay I'm still a little confused can you help me with one?

Answer 60

how far did you get

Answer 61

well I'm trying to figure out what to substitute

Answer 62

\[\Large \hat{p} = \frac{x}{n} = \frac{408}{865} \approx 0.471676 \approx 0.472\]

x = number of successes
n = sample size

Answer 63

z = 1.96 approximately based off the 95% confidence level

Answer 64

since P(-1.96 < Z < 1.96) = 0.95 approximately

Answer 65

okay I get it. One second

Answer 66

i keep getting -0.025 for the lower

Answer 67

what are you getting for the value of \[\Large \sqrt{\frac{\hat{p}*(1-\hat{p})}{n}}\]

Answer 68

0.017

Answer 69

\[\Large L = \hat{p} - z*\sqrt{\frac{\hat{p}*(1-\hat{p})}{n}}\]
\[\Large L = 0.472 - 1.96*\sqrt{\frac{0.472*(1-0.472)}{865}}\]
\[\Large L = 0.472 - 1.96*0.017\]
\[\Large L = 0.43868\]

things will be a bit off due to rounding errors, but it's close enough

Answer 70

Oh I see. it was my order of operations. I subtracted first.

Answer 71

if i'm doing a problem and I get a z score higher than 3.9 does that mean I did it wrong?

Answer 72

I would have to see the full problem

Answer 73

Suppose the number of students in a class for the Business Statistics program at a University has a mean of 23 with a standard deviation of 4.3. If 15 classes are selected randomly, find the probability that the mean number of students is between 20 and 30.

0.9699
0.6824
0.8412
0.9966

Answer 74

what z-scores did you get?

Answer 75

-2.70 and 6.30

Answer 76

me too

Answer 77

so you need to compute

P(-2.70 < Z < 6.30)

Answer 78

P(-2.70 < Z < 6.30) = P(Z < 6.30) - P(Z < -2.70) = ?? - ??

Answer 79

unfortunately, the table you have doesn't go up to 6.30

Answer 80

so use something like wolfram alpha

http://www.wolframalpha.com/input/?i=P(-2.70%20%3C%20Z%20%3C%206.30)&t=crmtb01&f=rc

Answer 81

okay thanks.

Answer 82

Assume that a sample is used to estimate a population proportion p. Find the margin of error E that corresponds to the given statistics and confidence level. Round the margin of error to four decimal places.

95% confidence; the sample size is 10,000, of which 40% are successes

0.0110
0.0072
0.0096
0.0126

Answer 83

should I use the equation from earlier?

Answer 84

\[\Large E = z*\sqrt{\frac{\hat{p}*(1-\hat{p})}{n}}\]

Answer 85

z = 1.96 at the 95% CL

http://www.wolframalpha.com/input/?i=95%25+Confidence

\(\Large \hat{p} = 0.4\)

n = 10,000

Answer 86

i got 0.0096

Answer 87

I did also

Answer 88

Would I do this like the other?

In one region, the September energy consumption levels for single-family homes are found to be normally distributed with a mean of 1050 kWh and a standard deviation of 218 kWh. If 50 different homes are randomly selected, find the probability that their mean energy consumption level for September is greater than 1075 kWh.

0.4562
0.0438
0.2090
0.2910

Answer 89

what's the z-score?

Answer 90

i got 0.81

Answer 91

good

Answer 92

so you need to compute P(Z > 0.81)

Answer 93

0.2090?

Answer 94

correct

Answer 95

is this like any of the others?

Answer 96

When 430 randomly selected light bulbs were tested in a laboratory, 224 lasted more than 500 hours. Find a point estimate of the proportion of all light bulbs that last more than 500 hours.

0.519
0.521
0.343
0.479

Answer 97

\[\Large \hat{p} = \frac{x}{n} = \frac{224}{430} \approx 0.52093 \approx 0.521\]

x = number of successes
n = sample size