Question
In: Statistics and Probability
The number of hours 10 students spent studying for a test and their scores on that...
The number of hours 10 students spent studying for a test and their scores on that test are shown in the table. Is there enough evidence to conclude that there is a significant linear correlation between the data? Use alphaequals0.05.
Hours, x= 0 1 2 4 4 5 5 6 7 8 Test score, y= 41 40 51 56 65 70 76 70 84 97
1. Setup the hypothesis for the test.
2. Identify the critical value(s). Select the correct choice below and fill in any answer boxes within your choice The critical value is ____ Or The critical values are -t0=____ &t0=____
3. Calculate the test statistic. T=_____(Round to three decimal places as needed)
4. What is your conclusion? There ____(is or is not) enough evidence at the 5% level of significance to conclude that there____ (is is not) a significant linear correlation between vehicle weight and variability in braking distance on a dry surface.
Solutions
Expert Solution
| X | Y | XY | X² | Y² |
| 0 | 41 | 0 | 0 | 1681 |
| 1 | 40 | 40 | 1 | 1600 |
| 2 | 51 | 102 | 4 | 2601 |
| 4 | 56 | 224 | 16 | 3136 |
| 4 | 65 | 260 | 16 | 4225 |
| 5 | 70 | 350 | 25 | 4900 |
| 5 | 76 | 380 | 25 | 5776 |
| 6 | 70 | 420 | 36 | 4900 |
| 7 | 84 | 588 | 49 | 7056 |
| 8 | 97 | 776 | 64 | 9409 |
| Ʃx = | 42 |
| Ʃy = | 650 |
| Ʃxy = | 3140 |
| Ʃx² = | 236 |
| Ʃy² = | 45284 |
| Sample size, n = | 10 |
| x̅ = Ʃx/n = 42/10 = | 4.2 |
| y̅ = Ʃy/n = 650/10 = | 65 |
| SSxx = Ʃx² - (Ʃx)²/n = 236 - (42)²/10 = | 59.6 |
| SSyy = Ʃy² - (Ʃy)²/n = 45284 - (650)²/10 = | 3034 |
| SSxy = Ʃxy - (Ʃx)(Ʃy)/n = 3140 - (42)(650)/10 = | 410 |
Slope, b = SSxy/SSxx = 410/59.6 = 6.8791946
y-intercept, a = y̅ -b* x̅ = 65 - (6.87919)*4.2 = 36.107383
Regression equation :
ŷ = 36.1074 + (6.8792) x
Sum of Square error, SSE = SSyy -SSxy²/SSxx = 3034 - (410)²/59.6 = 213.5302013
Standard error, se = √(SSE/(n-2)) = √(213.5302/(10-2)) = 5.16636
--
1.
Null and alternative hypothesis:
Ho: β₁ = 0
Ha: β₁ ≠ 0
2.
df = n-2 = 8
Critical value, t_c = T.INV.2T(0.05, 8) = 2.306
The critical values are -t0 = -2.306 & t0 = 2.306
3.
Test statistic:
t = b/(se/√SSxx) = 10.280
4.
Conclusion:
There is enough evidence at the 5% level of significance to conclude that there is a significant linear correlation between vehicle weight and variability in braking distance on a dry surface.
orchestra answered 1 year agoRelated Solutions
The number of hours 10 students spent studying for a test and their scores on that...
The number of hours 10 students spent studying for a test and
their scores on that test is represented in the table below.
Hours spent studying, x
0
2
4
5
5
5
6
7
8
9
Test scores, y
40
51
64
69
73
75
93
90
95
95
7) What amount of the variance in test score (y) do the hours
spent studying (x) account for this sample? (r2 =?) Interpret the
result in the context of the...
The number of hours 10 students spent studying for a test and their scores on that...
The number of hours 10 students spent studying for a test and
their scores on that test is represented in the table below.
Hours spent studying, x
0
2
4
5
5
5
6
7
8
9
Test scores, y
40
51
64
69
73
75
93
90
95
95
3) Describe the type of correlation and interpret the
correlation in the context of the data. Be specific in describing
the magnitude, direction, and strength of the relationship.
(12p)
4)...
The number of hours 10 students spent studying for a test and their scores on that...
The number of hours 10 students spent studying for a test and
their scores on that test are shown in the table. Is there enough
evidence to conclude that there is a significant linear correlation
between the data? Use alphaαequals= .01
X(hours) = top row; Y(score)=bottom row
0
1
2
3
4
4
5
6
7
8
41
46
52
50
65
68
75
85
82
91
Set up hypothesis test = ??
Critical Values = ??
Calculate test statistic;...
The number of hours 1010 students spent studying for a test and their scores on that...
The number of hours
1010
students spent studying for a test and their scores on that test
are shown in the table. Is there enough evidence to conclude that
there is a significant linear correlation between the data?
Use
alphaαequals=0.050.05.
Hours, x
00
11
22
44
44
55
55
66
77
88
Test score, y
4141
4040
5555
5656
6464
6666
7575
7878
8181
9797
LOADING...
Click here to view a table of critical values for Student's
t-distribution.
Setup...
The following data gives the number of hours 10 students spent studying and their corresponding grades...
The following data gives the number of hours 10 students spent
studying and their corresponding grades on their midterm exams.
Hours Spent Studying
0
0.5
1
2
2.5
3
4
4.5
5
5.5
Midterm Grades
60
63
75
81
84
87
90
93
96
99
Determine if r is statistically significant at the 0.01
level.
The following data gives the number of hours 10 students spent studying and their corresponding grades...
The following data gives the number of hours 10 students spent
studying and their corresponding grades on their midterm exams.
Hours Spent Studying 0 0.5 1 2 2.5 3.5 4 5 5.5 6
Midterm Grades 63 72 75 78 81 84 87 90 93 96
Step 1 of 3: Draw a scatter plot of the given data
Step 2 of 3: Estimate the correlation in words: positive,
negative, no correlation.
Step 3 of 3: Calculate the correlation coefficient, r. Round...
The following data give the number of hours 10 students spent studying and their corresponding grades...
The following data
give the number of hours 10 students spent studying and their
corresponding grades on their midterm exams.
Hours Studying
0
0.5
1
1.5
3
3
3.5
4
4
5.5
Midterm Grades
63
63
67
69
71
72
73
73
81
89
A. Calculate the sum
of squared errors (SSE). Use the values b0=61.3521 and b1=4.1338
for the calculations. Round your answer to three decimal
places.
B. Calculate the
estimated variance of errors, s2e. Round your answer to...
The number of hours spent studying by students on a large campus in the week before...
The number of hours spent studying by students on a large campus
in the week before the final exams follows a normal distribution
with standard deviation of 8.4 hours. A random sample of these
students is taken to estimate the population mean number of hours
studying.
a. How large a sample is needed to ensure that the probability
that the sample mean differs from the population mean by more than
2.0 hours is less than 0.05?
b. Without doing the...
The following data give the number of hours 10 students spent studying and their corresponding grades on their midterm exams
The following data give the number of hours 10 students spent studying and their corresponding grades on their midterm examsStep 1 of 5:Calculate the sum of squared errors (SSE). Use the values
b0=58.4077b0=58.4077 and b1=6.6860b1=6.6860 for the calculations.
Round your answer to three decimal places.Step 2 of 5:Calculate the estimated variance of errors, s2ese2. Round your
answer to three decimal places.Step 3 of 5:Calculate the estimated variance of slope, s2b1sb12. Round your
answer to three decimal places.Step 4 of 5:Construct...
The following data give the number of hours 5 students spent studying and their corresponding grades...
The following data give the number of hours 5 students spent
studying and their corresponding grades on their midterm exams.
Hours Studying
1
1
2
3
6
Midterm Grades
65
73
74
86
91
Step 1 of 5:
Calculate the sum of squared errors (SSE). Use the values
b0=65.9185 and b1=4.5698 for the calculations. Round your answer to
three decimal places.
Step 2 of 5:
Calculate the estimated variance of errors, s^2e. Round your
answer to three decimal places.
Step...
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