It is estimated that 35% of households own a riding lawn mower. A sample 19 households is studied. What is the mean number of ho
1 answer:
Answer:
The mean number of households who own a riding lawn mower is 6.65.
Step-by-step explanation:
For each household, there are only two possible outcomes. Either there is a riding lawn mower, or there is not. The probability of a household having a riding lawn mower is independent of other households. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
Probability of exactly x sucesses on n repeated trials, with p probability.
The expected value of the binomial distribution is:
35% of households own a riding lawn mower.
This means that
A sample 19 households is studied.
This means that
What is the mean number of households who own a riding lawn mower
The mean number of households who own a riding lawn mower is 6.65.
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Answer:
(a) The test statistic value is -4.123.
(b) The critical values of <em>t</em> are ± 2.052.
Step-by-step explanation:
In this case we need to determine whether there is evidence of a difference in the mean waiting time between the two branches.
The hypothesis can be defined as follows:
<em>H₀</em>: There is no difference in the mean waiting time between the two branches, i.e. <em>μ</em>₁ - <em>μ</em>₂ = 0.
<em>Hₐ</em>: There is a difference in the mean waiting time between the two branches, i.e. <em>μ</em>₁ - <em>μ</em>₂ ≠ 0.
The data collected for 15 randomly selected customers, from bank 1 is:
S = {4.21, 5.55, 3.02, 5.13, 4.77, 2.34, 3.54, 3.20, 4.50, 6.10, 0.38, 5.12, 6.46, 6.19, 3.79}
Compute the sample mean and sample standard deviation for Bank 1 as follows:
The data collected for 15 randomly selected customers, from bank 2 is:
S = {9.66 , 5.90 , 8.02 , 5.79 , 8.73 , 3.82 , 8.01 , 8.35 , 10.49 , 6.68 , 5.64 , 4.08 , 6.17 , 9.91 , 5.47}
Compute the sample mean and sample standard deviation for Bank 2 as follows:
(a)
It is provided that the population variances are not equal. And since the value of population variances are not provided we will use a <em>t</em>-test for two means.
Compute the test statistic value as follows:
Thus, the test statistic value is -4.123.
(b)
The degrees of freedom of the test is:
Compute the critical value for <em>α</em> = 0.05 as follows:
*Use a <em>t</em>-table for the values.
Thus, the critical values of <em>t</em> are ± 2.052.
