Find the product. (4m^2 – 5)(4m^2 + 5)
1 answer:
You might be interested in
Answer:
The first option is correct. Option A is correct.
LCL = 0.270, and UCL = 0.397
80% Confidence interval = (0.270, 0.397)
Step-by-step explanation:
The data for Y and N for the 90 companies is attached to this solution provided.
Y represents companies with at least 1 bilingual operator and N represents companies with no bilingual operator.
The number of Y in the data = 30
Hence, sample proportion of companies with at least one bilingual operator = (30/90) = 0.3333
Confidence Interval for the population proportion is basically an interval of range of values where the true population proportion can be found with a certain level of confidence.
Mathematically,
Confidence Interval = (Sample proportion) ± (Margin of error)
Sample proportion = 0.3333
Margin of Error is the width of the confidence interval about the mean.
It is given mathematically as,
Margin of Error = (Critical value) × (standard Error)
Critical value at 80% confidence level for sample size of 90 is obtained from the z-tables.
Critical value = 1.280
Standard error of the mean = σₓ = √[p(1-p)/n]
p = sample proportion
n = sample size = 90
σₓ = √[0.3333×0.6667)/90] = 0.0496891568 = 0.04969
80% Confidence Interval = (Sample proportion) ± [(Critical value) × (standard Error)]
CI = 0.3333 ± (1.28 × 0.04969)
CI = 0.3333 ± 0.0636021207
80% CI = (0.2696978793, 0.3969021207)
80% Confidence interval = (0.270, 0.397)
Hope this Helps!!!
Answer:
Yes, since the test statistic value is greater than the critical value.
Step-by-step explanation:
Data given and notation
represent the mean for the downtown restaurant
represent the mean for the freeway restaurant
represent the sample standard deviation for downtown
represent the sample standard deviation for the freeway
sample size for the downtown restaurant
sample size for the freeway restaurant
t would represent the statistic (variable of interest)
significance level provided
Develop the null and alternative hypotheses for this study?
We need to conduct a hypothesis in order to check if the true mean of revenue for downtown is higher than for freeway restaurant, the system of hypothesis would be:
Null hypothesis:
Alternative hypothesis:
Since we don't know the population deviations for each group, for this case is better apply a t test to compare means, and the statistic is given by:
(1)
t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.
Determine the critical value.
Based on the significance level and
we can find the critical values from the t distribution dith degrees of freedom df=36+36-2=55+88-2=70, we are looking for values that accumulates 0.025 of the area on the right tail on the t distribution.
For this case the value is
Calculate the value of the test statistic for this hypothesis testing.
Since we have all the values we can replace in formula (1) like this:
What is the p-value for this hypothesis test?
Since is a right tailed test the p value would be:
Based on the p-value, what is your conclusion?
Comparing the p value with the significance level given we see that
so we can conclude that we reject the null hypothesis, and the true mean for the downtown revenue restaurant seems higher than the true mean revenue for the freeway restaurant.
The best option would be:
Yes, since the test statistic value is greater than the critical value.
Answer:
f(g(x)) = -1
Step-by-step explanation: