Question

how do you find the critical value and rejection region of a claim that states that 55% of people eat breakfast. 250 random sample, mean is 56.4 eat breakfast, level of sig = .01

Answer 1

hypothesis testing eh ...

Answer 2

yes : )

Answer 3

havent actually gotten to these in stats class yet, but it is aking to finding a confidence interval i think; the level of sig at .01 produces a critical zscor of 2.575 if i recall correctly

Answer 4

yes that is correct

Answer 5

since the means is at 56.4; n = 250; x = 250(.55); confidencce interval = 99% ...

i can ti83 it maybe ... but got no idea if thats a good answer :)

Answer 6

E = zscore(sqrt(pq/n)) ....

= 20.25512

mean - E = 36.1448....

Answer 7

36.1448.. < mean < 76.6551...

is the confidence interval ; maybe :)

Answer 8

it looks like I need to find the z value since the sample size is greater than 30 and I cannot use the t-distribution chart. I'm unsure how to do find the z

Answer 9

anything over 30 can use the z charts

Answer 10

I also need to find the standardized test statistic for this same problem

Answer 11

.5-(.01/2) = .495 in the z table

Answer 12

the inversenorm of (.495) is a z-score of 2.575

Answer 13

the mean of 56.4 is about 22.5%; if we take the confidence interval and convert it into % numbers... by /250. we should be able to determine if 55% is inside or outside the interval

Answer 14

i get an interval of abt. 14% to 31%; so im gonna assume the 55% is a bad claim

Answer 15

I agree

Answer 16

standardized test statistic eh .... any clue as to how to get that :)

Answer 17

standard deviation is unknown as far as i can tell

Answer 18

\[z = \cfrac{\bar x-\mu_{\bar x}}{^{'}sd\ ^{'}/\sqrt{n}}\]

Answer 19

thats what the book gives me for test statistics about population mean when n>30

Answer 20

i think thats all i can glean from this :) hope it helps

Answer 21

thanks : ) it did