Question

Use the following to answer Question 9-10.The Lightning Bolt Electric Car company claims their new car “The Buzzer” can travel on average 170 miles on one charge (𝜇 = 170 𝑚𝑖𝑙𝑒𝑠). It is known that the population distribution associated with the distance per charge in normally distributed with standard deviation of 30 miles.
9. Approximately, what is the percentage of cars that travelled between 170 and 260 miles on average?
a. 95%
b. 47.5%
c. 49.85%

Answer 1

@jim_thompson5910

Answer 2

So for this one I used the standard normal distribution

Answer 3

P(170<=X<=260)

Answer 4

hint: empirical rule

Answer 5

okay so how do I know if i use 1std, 2st, or 3std?

Answer 6

what is the z score for x = 260 ?

Answer 7

Let me think about that for a min

Answer 8

\[\Large z = \frac{x - \mu}{\sigma}\]

Answer 9

it's 3 using z = (x-u)/sigma

Answer 10

we are looking at the 99% percentile. I am doing the work now

Answer 11

not quite 99% but close to it

Answer 12

80 for my lower bound and 260 for my upper bound...

Answer 13

empirical rule

http://www.nku.edu/~statistics/images/Using_1.gif

Answer 14

it owuld be 95% percentile

Answer 15

does that image help?

Answer 16

no those number at the base of the horizontal line confuse me lol the 80 is left of the mean correct?

Answer 17

we are going out 3std from the mean left and right so id say it captures quite a bit

Answer 18

80 is 3 std dev below the mean
260 is 3 std dev above the mean

99.7% of the population is within 3 std dev of the mean

Answer 19

how about this?

http://www.oswego.edu/~srp/stats/6895997.htm

Answer 20

much better your answer makes total sense but the key says other wise. they say it's 49.85% could be a typo

Answer 21

well you cut 99.7 in half to get 49.85

Answer 22

since we're going from 170 (midpoint) to 260

not from 80 to 260

Answer 23

okay so 170 was our mean? that makes more sense. they use mu = xbar?

Answer 24

mu = population mean
xbar = sample mean

they are related ideas but are different things

Answer 25

nvm nvm i just spotted why. Okay I am going to the next question.

Answer 26

11. An efficiency expert wishes to determine the average amount time that it takes to drill three holes in a certain metal clamp. How large of a sample will she need to be 95% confident that her sample average will be at most 15 seconds from the true mean? Assume σ=40.
a. 5
b. 6
c. 27
d. 28

Answer 27

lets see..

Answer 28

do i use the z score. The most difficult part is trying to figure out which formula to use.

Answer 29

hold on i think i got it

Answer 30

true mean = pop. mean?

Answer 31

\[\Large n = \left(\frac{z_c*\sigma}{E}\right)^2\]

Answer 32

let me find that in my lecture notes. I've seen that formula and have used it lots.

Answer 33

here's a video with an example

https://www.youtube.com/watch?v=bMPd9-XOLUQ

Answer 34

why would we not use z score ?

Answer 35

so the ME = marginal error would be 15 sec?

Answer 36

i got 27

Answer 37

yes, that's the error E

Answer 38

let me check

Answer 39

you should get 27.3180444444446

then you round up to get 28

why am I rounding up? because rounding to 27 makes the sample size too small

if n = 27 then the margin of error would be too big (it would be larger than 15)

Answer 40

as n increases, the margin of error decreases

Answer 41

okay

Answer 42

Use the following to answer Questions 12-15.
Bags of a certain brand of tortilla chips claim to have a net weight of 16 ounces. A consumer advocate group wishes to see if there is evidence that the mean net weight is less than advertised and so intends to test the hypotheses H0: 𝜇 = 16, Ha: 𝜇 < 16. A simple random sample of 24 bags has a sample mean of
̅𝒙 = 15.68 and the sample standard deviation to be s = 0.64.

12. In this case, which test is appropriate to test the given null and alternative hypotheses? a. two-sample t test
b. one sample t test
c. paired t test
d. pooled t test

13. The 99% Confidence Interval for the average net weight is:
a. (14.31 oz.,17.67 oz.)
b. (15.31 oz.,16.05 oz)
c. (15.41oz.,15.95 oz.)
d. (15.46 oz.,15.90 oz.)

14. Perform the test. Based on these data,
a. we would reject H0 at significance level 0.10 but not at 0.05.
b. we would reject H0 at significance level 0.05 but not at 0.020.
c. we would reject H0 at significance level 0.025 but not at 0.01.
d. we would reject H0 at significance level 0.01.

15. Suppose we were not sure if the population distribution of net weights was Normal. In which of the following circumstances would conditions of the t-procedure be violated? a. The mean and median of the data are nearly equal.
b. A stem and leaf plot of the data shows moderate outlier.
c. A histogram of the data shows extreme skewness.
d. The sample standard deviation is large.

Answer 43

thinking

Answer 44

one sample t test

Answer 45

yes for #12

Answer 46

i am doing 13 now.

Answer 47

b? my numbers are not super close to any of the above

Answer 48

(15.208, 16.151)

Answer 49

n = 16 right S/sqrt(16)

Answer 50

use a t table and tell me what the t critical value will be

Answer 51

2.947

Answer 52

incorrect

Answer 53

n =24
df = n-1 = 24-1 = 23

Answer 54

crap...

Answer 55

look at this table

http://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf

look in the df = 23 row
look at the 99% confidence column

Answer 56

the confidence levels are at the bottom

Answer 57

the n = 24 i used 16

Answer 58

16 is the mean not the sample size

Answer 59

2.807

Answer 60

yep

Answer 61

(15.313, 16.046)

Answer 62

B is the answer

Answer 63

tc = 2.807
s = 0.64
xbar = 15.68

use these formulas

L = xbar - tc*s/sqrt(n)
U = xbar + tc*s/sqrt(n)

Answer 64

yep I used those formulas. How would i go about answering 14. I feel comfortable answering 14 if i knew the probability but how would you answer knowing a confidence interval?

Answer 65

15.68 - 2.807*0.64/sqrt(24) = 15.3132950555738 --> 15.31

15.68 + 2.807*0.64/sqrt(24) = 16.0467049444262 --> 16.05

so yeah I'm getting (15.31, 16.05) too

Answer 66

d is the answer i believe because we only have 0.005 % in each tail

Answer 67

0.05

Answer 68

you'll have to set up confidence intervals for the other confidence levels (based on the significance levels)

example: alpha = 0.01 ---> confidence level = 99%

Answer 69

okay hold on

Answer 70

answer is d
a. 90%
b. 99.5%
c. 75%
d. 99%

Answer 71

i am wrong... obviously i have no clue what is going on here..

Answer 72

why would we reject the H0 at a significance level 0.025

Answer 73

maybe it will help to do a regular t test

Answer 74

what is the t test statistic in this case?

Answer 75

with df = 23?

Answer 76

the df doesn't factor into finding the test statistic

Answer 77

t test statistic

\[\LARGE t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}}\]

Answer 78

give me a sec

Answer 79

I used Xbar = 15.68, S = 0.64, n = 24 , u = 16 and i got -2.449

Answer 80

now use this calculator

http://stattrek.com/online-calculator/t-distribution.aspx

to find the p value

Answer 81

df = 23
t score = -2.45 approx

Answer 82

okay if i don't respond right away is because my inteernet conection is flipping out

Answer 83

0.9888

Answer 84

that probability is quite large why would we reject?

Answer 85

see attached

Answer 86

nvm i didn' use a negative sign... 0.0112

Answer 87

we would reject at significance level 0.025 but not at 0.01? That doesn't make sense to me...

Answer 88

0.025 is a probability?

Answer 89

what does the number represent?

Answer 90

the value 0.0112 is the area under the curve to the left of t = -2.45

it is the p value in this case

Answer 91

now recall that

if `pvalue < alpha`, you reject the null hypothesis

Answer 92

ohhh okay... significance = alpha

Answer 93

yes

Answer 94

hmmm thinking... i am on 15

Answer 95

would it be c because a t distribution needs to be symmetric about the mean?

Answer 96

yay i am right lol

Answer 97

I checked the key.