1) A study was done on body temperature of men in the table. Assume that the samples are independent simple random samples selected from normally distributed populations, and do not assume that population deviation are equal. Complete (a) and (b) below.
Men Women
u U 1 U2
n 11 59
x 97.53 97.22
s 0.94 0.66
A) What are the null hypotheses?
Null Hypothesis: H0: µ1 = µ2
Alternate Hypothesis: H1: µ1 > µ2
Let
1.047
df = 11-1=10
P Value = 0.15987
Since p-value is greater than 0.05 so we fail to reject null hypothesis. Thus, there is enough evidence to support that men have higher mean body temperature than woman.
B) Construct a confidence interval suitable for testing claim that men have higher mean body temperature than woman.
df = 11-1=10
critical t-value = 1.81
We will use one sided confidence interval,
Limit will be (97.53-97.22)+ 1.81* = 0.846
2) Listed below are ages of actress and actors at the time they won an award for categories of best actress and best actor. Use the sample data to test for a difference between the ages of actresses and actors when they won the award. Use a 0.01 significance level. Assume that the paired sample data is simple random sample and that the differences have distribution that is approximately normal.
Actress’s age 16 26 20 53 22
Actor’s age 45 41 62 55 40
What are the null hypotheses for the hypothesis test ?
Solution:
Null Hypothesis: H0: µd = 0
Alternate Hypothesis: H1: µd ≠ 0
Identify the test statistics ?
We will use t-statistic.
Actress’s age / Actor’s age / Difference16 / 45 / -29
26 / 41 / -15
20 / 62 / -42
53 / 55 / -2
22 / 40 / -18
Mean difference / -21.2
Standard deviation (difference) / 15.09
P Value=0.0348
Since p-value is greater than 0.01 so we fail to reject null hypothesis. Thus there is no difference between their agaes.
3) The data below are yields for two different types of corn seed land. Assume that the data are simple random samples and that the differences have a distribution that is approximately normal. Construct a 95% confidence interval estimate of the difference between type 1 and type 2 yields.
What does the confidence interval suggest about farmer Joe’s claim that type I seed is better than type 2 seed?
Type 1 2140 2086 2055 2565 2136 1981 2155 1530
Type 2 2045 1999 2047 2486 2152 1984 2129 1425
In this example u d is the mean value of the differences d for the population of all pairs of data, where individual difference d can is defined as type 1 seed yield minus the type 2 seed yield.
The 95% confidence interval is ______< ud < ______
Solution:
Type 1 / Type 2 / Difference2140 / 2045 / 95
2086 / 1999 / 87
2055 / 2047 / 8
2565 / 2486 / 79
2136 / 2152 / -16
1981 / 1984 / -3
2155 / 2129 / 26
1530 / 1425 / 105
Mean difference / 47.625
Standard deviation (difference) / 48.873
df = 8-1 = 7
Critical t-value = 2.365
95% confidence interval is
95% confidence interval is 6.76 < ud < 88.49.
4) Use the given data to complete (a) through (c)below.
X 10 8 13 9 11 14 6 4 12 7 5
Y 9.15 8.14 8.75 8.78 9.26 8.09 6.13 3.09 9.14 7.26 4.74
A) Construct a scatter plot?
Solution:
B) Find the linear correlation coefficient r, then determine whether there is sufficient evidence to support the claim of linear correlation between the two variables?
Solution:
X / Y / XY / X^2 / Y^210 / 9.15 / 91.5 / 100 / 83.7225
8 / 8.14 / 65.12 / 64 / 66.2596
13 / 8.75 / 113.75 / 169 / 76.5625
9 / 8.78 / 79.02 / 81 / 77.0884
11 / 9.26 / 101.86 / 121 / 85.7476
14 / 8.09 / 113.26 / 196 / 65.4481
6 / 6.13 / 36.78 / 36 / 37.5769
4 / 3.09 / 12.36 / 16 / 9.5481
12 / 9.14 / 109.68 / 144 / 83.5396
7 / 7.26 / 50.82 / 49 / 52.7076
5 / 4.74 / 23.7 / 25 / 22.4676
SUM / 99 / 82.53 / 797.85 / 1001 / 660.6685
Correlation(r) = NΣXY - (ΣX)(ΣY) / Sqrt([NΣX2- (ΣX)2][NΣY2- (ΣY)2])
r = (11*797.85 – 99*82.53)/sqrt((11*1001-99^2)*(11*660.6685-82.53^2)
r = 0.8155
n = 11
Critical value = ±0.602
Therefore r = 0.816 indicates a significant (positive) linear correlation. Yes; there is sufficient evidence to support the claim that there is a linear correlation between the 2 variables.
C) Identify the feature of the data that would be missed if part (b) was completed without constructing the scatter plot ?
The scatterplot indicates that the relationship between the variables is quadratic, not linear.
5) The data show the time intervals after an eruption (to the next eruption) of a certain geyser. Find the regression equation, letting the first variable be the independent (x) variable. Find the best predicted time of the interval after an eruption given that the current eruption has a height of 98 feet. Use a significance level 0.05.
Height (ft) 71 57 73 61 91 57 96 87
Interval after (min) 71 68 67 65 81 64 72 76
Solution:
X / Y / XY / X^2 / Y^271 / 71 / 5041 / 5041 / 5041
57 / 68 / 3876 / 3249 / 4624
73 / 67 / 4891 / 5329 / 4489
61 / 65 / 3965 / 3721 / 4225
91 / 81 / 7371 / 8281 / 6561
57 / 64 / 3648 / 3249 / 4096
96 / 72 / 6912 / 9216 / 5184
87 / 76 / 6612 / 7569 / 5776
SUM=593 / SUM=564 / SUM=42316 / SUM=45655 / SUM=39996
Slope(b) = (NΣXY - (ΣX)(ΣY)) / (NΣX2- (ΣX)2)
b = (8*42316-593*564)/(8*45655-593^2) = 0.2999
Intercept(a) = (ΣY - b(ΣX)) / N
a = (564-0.2999*593)/8 = 48.2699
Regression Equation(Y) = a + bX
Y = 48.2699 + 0.2999X
Put X = 98 feet
Y = 48.2699 + 0.2999*98 = 77.66 min.