Answer:
Part A: yes
Part B: 50%
Step-by-step explanation:
There are two ways we can use a normal approximation for a sample: if the population it is from is normally distributed, or if the sample is sufficiently large (n > 30). In this case, we don't know if the population is normally distributed, but we do know the sample size is larger than 30, so we can use normal approximation.
The sample proportion is normally distributed, with a mean equal to the mean of the population. So there is a 50% chance that sample proportion is less than the population proportion.
Divide 100 by 8: 100/8=12.5. 12 full rows and 1 half row so the answer is 12 :)
Answer:
we reject H0 and conclude that management's belief about variance is untrue.
Step-by-step explanation:
H0 : σ² = 250
H1 : σ² ≠ 250
Sample size, n = 25
xbar = 50.6 ; s² = 500
α = 0.01 ;
The test statistic :
χ² = [(n - 1)s²] ÷ σ²
χ² = [(25 - 1)* 500] ÷ 250
χ² = (24 * 500) / 250
χ² = 12000 / 250
χ² = 48
The critical value:
df = n-1 ; 25 - 1 = 24
χ²(0.01/2 ; 24) = 45.559
Decison :
Reject H0 : if |χ² statistic| > critical value
Since 48 > 45.559
Hemce, we reject H0 and conclude that management's belief about variance is untrue.
