Question

The mean age of De Anza College students in a previous term was 26.6 years old. An instructor thinks the mean age for online students is older than 26.6. She randomly surveys 56 online students and finds that the sample mean the 5% level s 29.8 with a standard deviation of 2.1. Conduct a hypothesis test Note: If you are using a Student's t-distribution for the problem, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.) Part (a) Part (b) Part (c Part (d) | Part (e three decimal places.) What is the test statistic? (If using the z distribution round your answers to two decimal places, and if using the t distribution round your answers Part (f) Part (g) Part (h) Part () Construct a 95% confidence interval for the true mean. Sketch the graph of the situation. Label the point estimate and the lower and upper bounds of the confidence interval. (Round your answers to two decimal places.) 95% CI. 29.8 Bloomberg Businessweek, New York City's most recent adult smoking rate 14%. Suppose that a survey is conducted to determine this year's rate. Eight out of 70 randomly According to an article determine if the rate is still 14% or if it has decreased. chosen N. Y. City residents reply that they smoke. Conduct a hypothesis test at the 5 % level t normally distributed. (In general, you must first prove that assumption, though.) Note: If you are using Student's t-distribution for the problem, you may assume that the underlying population Part (a) Part (b) Part (c) EPart (d) State the distribution to use for the test. (Round your answers to four decimal places.) PN EPart (e) What is the test statistic? (If using the z distribution round your answers to two decimal places, and if using the t distribution round your answers to three decimal places.) O Part (f) four decimal places. What is the p-value? (Round your answert

Answer 1

We are given ,

µ = 26.6 , $\bar{x}$ = 29.8 , s = 2.1 and n = 56

Part e) Test statistic :

t =
X n s/ S
=
29.8-26.6 2.1/V56
  

t = 11.403

Part i) 95% confidence interval

Lower = $\bar{x}$ - E

Upper = $\bar{x}$ + E

E is margin of error =    ; t is critical value follows t distribution with degrees of freedom (df) = n-1

t * Vn

We are given confidence level = 0.95

We can find critical value using excel function =TINV( α , d.f )

α = 1 - confidence level = 1 - 0.95 = 0.05

degrees of freedom (d.f) = n - 1 = 56-1 = 55

t =TINV(0.05,55)

t = 2.004

E is margin of error =
t * Vn
=  
2.004 2.1 56

E = 0.5624

Lower = $\bar{x}$ - E = 29.8 - 0.5624

Lower = 29.24

Upper = $\bar{x}$ + E = 29.8 + 0.5624

Upper = 30.36

29.24 30.36

#2) We are given x = 8 and n = 70 and P = 0.14

Part d) According to sampling distribution of proportion , sample proportion $\hat{p}$ follows normal distribution with mean = P = 0.14 and standard deviation =
Р* (1- Р)
=
0.14 (1 0.14) 70
= 0.0415

P' - N ( 0.14 , 0.0415 )

Part e)  

Test statistic : We have $\hat{p}$ = x/n = 8/70 = 0.1143

Z =
P- P P(1-P) n
=  
0.1143 0.14 0.0415

Z = -0.62

Part f)

This is left tail test , therefore p value would be the are to the left of -0.62

P value = P ( z ≤ -0.62 )

P- value = 0.2676