Answer:
proportion of gamers who prefer console does not differ from 29%
Step-by-step explanation:
Given :
n = 341 ; x = 95 ; Phat = x / n = 95/341 = 0.279
H0 : p = 0.29
H1 : p ≠ 0.29
The test statistic :
T = (phat - p) ÷ √[(p(1 - p)) / n]
T = (0.279 - 0.29) ÷ √[(0.29(1 - 0.29)) / 341]
T = (-0.011) ÷ √[(0.29 * 0.71) / 341]
T = -0.011 ÷ 0.0245725
T = - 0.4476532
Using the Pvalue calculator from test statistic score :
df = 341 - 1 = 340
Pvalue(-0.447, 340) ; two tailed = 0.654
At α = 0.01
Pvalue > α ; We fail to reject the null and conclude that there is no significant evidence that proportion of gamers who prefer console differs from 29%
Given:
n = 20, sample size
xbar = 17.5, sample mean
s = 3.8, sample standard deiation
99% confidence interval
The degrees of freedom is
df = n-1 = 19
We do not know the population standard deviation, so we should determine t* that corresponds to df = 19.
From a one-tailed distribution, 99% CI means using a p-value of 0.005.
Obtain
t* = 2.8609.
The 99% confidence interval is
xbar +/- t*(s/√n)
t*(s/√n) = 2.8609*(3.8/√20) = 2.4309
The 99% confidence interval is
(17.5 - 2.4309, 17.5 + 2.4309) = (15.069, 19.931)
Answer: The 99% confidence interval is (15.07, 19.93)
Answer:
First multiply 30 by 3 = 90 then subtract 90 from 156 to get 66
Step-by-step explanation:
66 would be your thirtieth term.
The correct answer is red.
