Statistics Homework Questions

The assignment content involves solving various statistical problems. Question 1 deals with finding mean amount of work per week for three randomly selected professors. Question 2 is about calculating the confidence interval estimate of population mean. Question 3 is related to finding the confidence interval estimate of population mean and testing whether the sample mean defer from a given value. Question 4 is about performing a two-tail test to see if the sample means are equal or not. Question 5 involves solving an F-test problem to determine if treatment means are equal or not. Finally, Question 6 deals with linear regression analysis and performs various tests such as coefficient of determination, t-test for slope, etc.

Individual Assignment
Question 1
Part a)
i)
P ( Z >−0.8 )=0.788
ii)
P (−1.3< Z←0.7 )=0.145
iii)
Z0.2=0. 5793
Part b)
Z= x −μ
σ
¿ x−65
4
i)
P ( x>70 )=P (Z> 70−65
4 )
¿ P ( Z >1.25 )
¿ 1−0.8944
¿ 0.1056
ii)
P ( x< 60 ) =P ( Z < 60−65
4 )
¿ P ( Z ←1.25 )
¿ 1−0.8944

¿ 0.1056
iii)
P ( 55< x< 70 )=P ( 55−65
4 <Z <70−65
4 )
¿ P (−2.5<Z<1.25 )
¿ 0.8944−0.0062
¿ 0.8882
Question 2
Part a)
i)
Here , μ=100∧σ=20
Therefore , z= 96−100
20 =−0.2
P ( X <96 )=P ( z←0.2 )
¿ 0.4207
ii)
P ( 96< X< 105 ) =P ( 96−100
20 < z< 105−100
20 )
¿ P (−0.2< z <0.25 )
¿ 0.178

Part b)
Z= x −μ
σ
¿ x−52
6
i)
P ( x> 60 )=P (Z >60−52
6 )
¿ P ( Z >1.33 )
¿ 0.0918
ii)
Let X̅ denote the mean amount of work per week for three randomly selected professors.
Since, X N ( 52 ,62 )
P ( X >60 ) =P (Z> 60−52
√ 12 )
¿ P ( Z >2.31 )
¿ 1−0.9896
¿ 0.0104

Question 3
Part a)
Given X =500 , σ=12 , n=50
i)
α=0.05
α
2 =0.025
Z α
2
=1.96
Confidence interval estimate of population mean
X −
Z α
2
∗σ
√ n , X +
Z α
2
∗σ
√ n
500− 1.96∗12
√50 , 500+ 1.96∗12
√50
496.674 ,503.326
The lower and upper confidence limits are LCL = 496.674 and UCL =503.326 respectively.
ii)
α=0.01
α
2 =0.005
Z α
2
=2.575
Confidence interval estimate of population mean
X −
Z α
2
∗σ
√ n , X +
Z α
2
∗σ
√ n

500− 2.575∗12
√50 , 500+ 2.575∗12
√50
495.63 , 504.37
The lower and upper confidence limits are LCL = 495.63 and UCL =504.37 respectively.
Part b)
Given X =120 , s=15 , n=51
i)
α=0.05
α
2 =0.025
Z α
2
=1.96
Confidence interval estimate of population mean
X −
Z α
2
∗s
√ n , X+
Z α
2
∗s
√ n
120− 1.96∗15
√51 , 120+ 1.96∗15
√51
115.883 , 124.117
The lower and upper confidence limits are LCL = 115.883 and UCL =124.117 respectively.
ii)
α=0.01
α
2 =0.005
Z α
2
=2.575
Confidence interval estimate of population mean
X −
Z α
2
∗σ
√ n , X +
Z α
2
∗σ
√ n

120− 2.575∗15
√51 , 120+ 2.575∗15
√51
114.592 ,125.408
The lower and upper confidence limits are LCL = 114.592 and UCL =125.408 respectively.
Question 4
Part a)
The null and alternative hypotheses are given as:
H0: μ = 10
H1: μ ≠ 10
At the centre of the hypothesized distribution will be the highest possible value for which H0
could be true, μ0 = 10.
The significance level as given will be α = 0.05
The population standard deviation (σ) is known and the sample size is large, so the normal
distribution is appropriate and the test statistics (ᴢ) is calculated as:
z = (x̅ - μ0)/σ
= (10 – 10)/ (10)
= 0
The calculated value, z = 0 at the 0.05 level of significance, so the null hypothesis is not rejected.
Therefore, the results suggest that the mean does not defer from 10.
Part b)
Let us formulate the null and alternative hypotheses.
H0: μ = 180
H1: μ ≠ 180

At the centre of the hypothesized distribution will be the highest possible value for which H0
could be true, μ0 = 180.
The significance level as given will be α = 0.05
The population standard deviation (σ) is known and the sample size is large, so the normal
distribution is appropriate and the test statistics (ᴢ) is calculated as:
z = (x̅ - μ0)/σ
= (175 – 180)/ (22/√200)
= -3.214
For a two-tail z-test in which α = 0.05, z = -1.96 and z = +1.96 will be the respective boundaries
for lower and upper tails of 0.025 each. These are the critical values for the test and they identify
the non rejection and rejection regions i.e. reject H0 if calculated z < -1.96 or z > +1.96,
otherwise do not reject.
The calculated value, z = -3.214 falls within the rejection region of the diagram. So at the 0.05
level of significance, the null hypothesis is rejected.
The results suggest that the population mean defers from 180.
P-value = (-3.214≤Z ≤ 3.214) = 0.00132 which is significant at p < 0.05.

Question 5
Given in the incomplete ANOVA table:
SS total = 1555.57
SS (for treatment) = 583.39
SS (for error) = 972.18
The null and the alternative hypotheses are:
H0:
1 =
2 =
3 =
4=
5
H1: Treatment means are not all equal.
Since the population distributions are approximately normal, the distribution of the test statistic
is approximately F-distribution, with df = (k – 1, n – k) = (4, 50).
For df = (4, 50), F = F0.05 = 2.557. {From F – distribution table}
The decision rule is: reject H0 in favour of H1 if the computed F-value is greater than 2.782.
MSE = SSE / df (for error)
= 972.18 / 50
Hence, MSE = 19.44.
MST = SST / df (for treatment)
= 583.39/4 = 145.85
F = MST / MSE
= 145.85 / 19.44
= 7.5
This gives us the following complete ANOVA table:
Source of Variation SS df MS F
Treatment 583.39 4 145.85 7.5
Error 972.18 50 19.44
Total 1555.57 54

Since the computed F-value (=7.5) is more than 2.557, we conclude that we have sufficient
evidence, at  = 0.05, to reject H0, that is, to infer that the treatment means are not all equal.
Question 6
a)
Intercept coefficient ( A ) =Standard Erroe∗t Stat
¿ 6748∗7.035
¿ 47472.18
t Stat Age ( B )= Age Coefficient
Standard Error
¿−2658
856
¿−3.105
b)
The slope of the regression line is negative (-2658). This means that when the independent
variable (age) increases, the dependent variable (price) decreases and vice-versa.
c)
The estimated linear regression line is:
^y=−2658 x + 47472.18
d)
coefficient of determination=R−square
¿ 0.077
The coefficient of determination tells us that 7.7% of the variation in price of used cars in dollars
is accounted for variation in age of the cars in years. Since this value is a very small, this means

that the data does not form a perfect line. Instead the data are scattered around the best fitting,
linear regression line and there will be error in predictions.
e)
coefficient of correlation=R
¿ √0.077
¿−0.2775(slopeis negative)
The coefficient of correlation is negative. This tells us that the dependent (price) and independent
variable (age) have a linear, weak and negative relationship.
f)
The t-test for the slope (b1) of the regression line:
 Null hypothesis H0 : b1 ≤ 0
 Alternative hypothesis H1 : b1 > 0
 Test statistics
t = (b1 - 0)/sb1
where b1 = slope of the regression line
sb1 = the estimated standard deviation of the slope
Testing H0: b1 ≤ 0 versus H1: b1 > 0
The observed value of the test statistic is
Observed t = (b1 - 0)/sb1
= (-2658 – 0)/856
= -3.105
Referring to the t-distribution table, for the 0.05 level of significance and number of
degrees of freedom (df) = n – 2 = 117 – 2 = 115, the critical values of t are – 1.981 to +
1.981.
Thus, at the 0.05 level, the test statistic is outside the critical values and so we are able to
reject the null hypothesis that the slope of the regression line could be negative.
So the result suggests that the slope is positive.
g)
price=−2658∗3+ 47472.18

¿ $ 39498.18