Answer:
Step-by-step explanation:
1)
Mean = (112 + 120 + 98 + 55 + 71 + 35 + 99 + 124 + 64 + 150 + 150 + 55 + 100 + 132 + 20 + 70 + 93)/17 = 91.1
Standard deviation = √(summation(x - mean)/n
Summation(x - mean) = (112 - 91.1)^2 + (120 - 91.1)^2 + (98 - 91.1)^2 + (55 - 91.1)^2 + (71 - 91.1)^2 + (35 - 91.1)^2 + (99 - 91.1)^2 + (124 - 91.1)^2 + (64 - 91.1)^2 + (150 - 91.1)^2 + (150 - 91.1)^2 + (55 - 91.1)^2 + (100 - 91.1)^2 + (132 - 91.1)^2 + (20 - 91.1)^2 + (70 - 91.1)^2 + (93 - 91.1)^2 = 23550.97
s = √23550.97/17 = 37.2
Confidence interval is written in the form,
(Sample mean - margin of error, sample mean + margin of error)
The sample mean, x is the point estimate for the population mean.
Margin of error = z × s/√n
Where
s = sample standard deviation = 37.2
n = number of samples = 17
From the information given, the population standard deviation is unknown and the sample size is small, hence, we would use the t distribution to find the z score
In order to use the t distribution, we would determine the degree of freedom, df for the sample.
df = n - 1 = 17 - 1 = 16
Since confidence level = 95% = 0.95, α = 1 - CL = 1 – 0.95 = 0.05
α/2 = 0.05/2 = 0.025
the area to the right of z0.025 is 0.025 and the area to the left of z0.025 is 1 - 0.025 = 0.975
Looking at the t distribution table,
z = 2.12
Margin of error = 2.12 × 37.2/√17
= 19.13
The 95% confidence interval for the hemoglobin reading for all the patients in the hospital is
91.1 ± 19.13
2) Confidence interval for population proportion is written as
Sample proportion ± margin of error
Margin of error = z × √pq/n
Where
z represents the z score corresponding to the confidence level
p = sample proportion. It also means probability of success
q = probability of failure
q = 1 - p
p = x/n
Where
n represents the number of samples
x represents the number of success
From the information given,
n = 500
x = 60
p = 60/500 = 0.12
q = 1 - 0.12 = 0.88
To determine the z score, we subtract the confidence level from 100% to get α
α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05
This is the area in each tail. Since we want the area in the middle, it becomes
1 - 0.05 = 0.95
The z score corresponding to the area on the z table is 1.645. Thus, Tthe z score for a confidence level of 90% is 1.645
Therefore, the 90% confidence interval is
0.12 ± 1.645√(0.12)(0.88)/500
Confidence interval = 0.12 ± 0.024
3) We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean
For the null hypothesis,
µ = 19410
For the alternative hypothesis,
µ > 19410
This is a right tailed test.
Since the population standard deviation is given, z score would be determined from the normal distribution table. The formula is
z = (x - µ)/(σ/√n)
Where
x = sample mean
µ = population mean
σ = population standard deviation
n = number of samples
From the information given,
µ = 19410
x = 22098
σ = 6050
n = 40
z = (19410 - 22098)/(6050/√40) = - 2.81
Looking at the normal distribution table, the probability corresponding to the z score is 0.0025
p value = 0.0025
Since alpha, 0.01 > than the p value, 0.0025, then we would reject the null hypothesis. Therefore, At a 1% level of significance, there is sufficient evidence to conclude that the cost of attendance has increased.
