Respuesta :
Let [tex] x [/tex] be the number of students who eat cauliflower.
Therefore, [tex] x=4 [/tex].
Let [tex] n [/tex] be the total number of students surveyed.
Therefore, [tex] n=39 [/tex]
Thus, [tex] \hat p [/tex][tex] =\frac{4}{39} =0.10256 [/tex]
Now, for 90% confidence level, from the table we know that [tex] Z=1.645 [/tex].
The formula for the interval range of proportion of students is :
[tex] p= \hat p\pm Z\sqrt{\frac{\hat p(1-\hat p)}{n}} [/tex]
Plugging in the values we get:
[tex] p=0.10256\pm 1.645\sqrt{\frac{0.10256(1-0.10256)}{39}}=0.10256\pm 0.04858=0.15114, 0.05398 [/tex]
Thus, Jane is 90% confident that the population proportion p, for students who eat cauliflower in her campus is between 5.398% and 15.114% (after converting the answer we got to percentage).
The 90% confidence interval for the proportion of students who eat the food will be (0.15114, 0.05398).
How to compute the confidence interval?
The following can be deduced from the information:
number of students = 39
proportion = 4/39 = 0.10256
The 90% confidence interval in the table is 1.645.
Therefore, the lower limit for the interval range will be:
= 0.10256 - 1.645[(✓0.10256(1 - 0.10256)/39]
= 0.05398
The upper limit will be:
= 0.10256 + 1.645[(✓0.10256(1 - 0.10256)/39]
= 0.15114
Learn more about confidence interval on:
https://brainly.com/question/15712887
