Linda had an orange ribbon and a blue ribbon. The orange ribbon was 2 meters long. The blue ribbon was 4/5 as long as the orange
2 answers:
Answer:
A - 1.6 meter and B - 0.40 meter
Step-by-step explanation:
Given:
The orange ribbon was 2 meters long.
The blue ribbon was 4/5 as long as the orange ribbon.
Linda cut off a piece of the blue ribbon.
The length of the cut piece was 25% of the length of the blue ribbon.
To find:
A) What was the length of the blue ribbon before it was cut?
B) Find the length of the piece of the blue ribbon that Linda cut off.
Solution:
Length of orange ribbon = 2 m
Length of blue ribbon = (given)
As the length of the cut piece was 25% of the length of the blue ribbon,
Therefore the length of the cut piece =
= meter
Therefore, A : The length of the blue ribbon before it was cut is meter that is 1.6 meter
B : The length of the piece of the blue ribbon that Linda cut off is 0.40 meter
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Answer:
90% confidence interval is ( -149.114, -62.666 )
Step-by-step explanation:
Given the data in the question;
Sample 1 Sample 2
x"₁ = 259.23 x"₂ = 365.12
s₁ = 34.713 s₂ = 48.297
n₁ = 5 n₂ = 10
With 90% confidence interval for μ₁ - μ₂ { using equal variance assumption }
significance level ∝ = 1 - 90% = 1 - 0.90 = 0.1
Since we are to assume that variance are equal and they are know, we will use pooled variance;
Degree of freedom DF = n₁ + n₂ - 2 = 5 + 10 - 2 = 13
Now, pooled estimate of variance will be;
= [ ( n₁ - 1 )s₁² + ( n₂ - 1)s₂² ] / [ ( n₁ - 1 ) + ( n₂ - 1 ) ]
we substitute
= [ ( 5 - 1 )(34.713)² + ( 10 - 1)(48.297)² ] / [ ( 5 - 1 ) + ( 10 - 1 ) ]
= [ ( 4 × 1204.9923) + ( 9 × 2332.6 ) ] / [ 4 + 9 ]
= [ 4819.9692 + 20993.4 ] / [ 13 ]
= 25813.3692 / 13
= 1985.64378
Now the Standard Error will be;
= √[ (
/ n₁ ) + (
/ n₂ ) ]
we substitute
= √[ ( 1985.64378 / 5 ) + ( 1985.64378 / 10 ) ]
= √[ 397.128756 + 198.564378 ]
= √595.693134
= 24.4068
Critical Value = =
= 1.771 { t-table }
So,
Margin of Error E = × [ (
/ n₁ ) + (
/ n₂ ) ]
we substitute
Margin of Error E = 1.771 × 24.4068
Margin of Error E = 43.224
Point Estimate = x₁ - x₂ = 259.23 - 365.12 = -105.89
So, Limits of 90% CI will be; x₁ - x₂ ± E
Lower Limit = x₁ - x₂ - E = -105.89 - 43.224 = -149.114
Upper Limit = x₁ - x₂ - E = -105.89 + 43.224 = -62.666
Therefore, 90% confidence interval is ( -149.114, -62.666 )
Answer:
The approximate p-value for the suitable test
<em>0.05 < p < 0.1</em>
|t| = |-1.5806| = 1.5806
t = 1.5806 < 2.009 at 0.05 level of significance
<em>Carisoprodol, a generic muscle relaxer, claims to have, on average, is equal to 120 milligrams of active ingredient.</em>
<u>Step-by-step explanation</u>:
<u><em>Step(i)</em></u>:-
Given mean of the Population 'μ' = 120 milligrams
Given random sample size 'n' = 50
Given mean of the sample x⁻ = 116.2 milligrams
Standard deviation of the sample 'S' = 17 milligrams
Null hypothesis : 'μ' = 120
Alternative hypothesis : 'μ' < 120
<u><em>Step(ii):</em></u>-
<em>Test statistic </em>
<em>Degrees of freedom</em>
<em>ν = n-1 = 50 -1 =49</em>
t₀.₀₅ = 2.009
|t| = |-1.5806| = 1.5806
t = 1.5806 < 2.009 at 0.05 level of significance
<em>Null hypothesis is accepted</em>
<em>Carisoprodol, a generic muscle relaxer, claims to have, on average, is equal to 120 milligrams of active ingredient.</em>
<u><em>P- value:-</em></u>
<em>The test statistic |t| = 1.5806 at 49 degrees of freedom</em>
<em>The test statistic value is lies between 0.05 to 0.1</em>
<em>0.05 < p < 0.1</em>
<u><em></em></u>