May somebody help me with this please?
2 answers:
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Answer:
1. ( 19.1416, 23.5384,)
2. (0.276348, 0.46651)
3. the sample size = 170.73 approximately 171
4. sample size = 334.07 approximately 334
5. sample of 334 should be taken by manager
Step-by-step explanation:
mean = bar x = 21.34 dollars
size of sample n = 70
standard deviation of sample = 9.22
we use t distribution as the population standard deviation is not known.
95% Confidence interval
1-α = 0.95
α = 0.05
degree of freedom = 70-1 = 69
α/2 = 0.025
using the t distribution tsble,
= 1.9949
confidence interval =
= 21.34 +- (1.9949*1.10200)
= 21.34 + 2.1984, 21.34 - 2.1984
= (23.5384, 19.1416)
the confidence interval of the mean amount spent at the supply store can be written as 19.1416<u<23.5384
2. sixe of those who only have a cat
p = 26/70 = 0.371429
at 90 % confidence interval,
1-α = 0.90
α = 0.10
we use the z table here
z(0.10/2) = Z(0.05)
= 1.645
= 0.371429 +-( 1.645 x 0.0578)
= 0.371429 + 0.095081, 0.371429 - 0.095081
= (0.276348, 0.46651)
3. sd = 10$
margin of erro,r e = 1.50$
α = 0.05
using z table
α/2 = Z0.025
= 1.96
sample size = 1.96² * 10² / 1.50²
= 3.8416 * 100/ 2.25
= 170.73
the sample size is approximately 171
d. we have 0.5 as sample proportion now
margin of error = 0.045
α = 0.10
Zα/2 = 0.05
= 1.645
sample size = 1.645²x0.5(1-0.5) / 0.045²
= 0.676506/0.002025
= 334. 07
sample size = 334
5. sample of 334 should be taken by manager