A random sample of 20 recent weddings in a country yielded a mean wedding cost of $ 26,388.67. Assume that recent wedding costs in this country are normally distributed with a standard deviation of $8200.Complete parts (a) through (c) below. a. Determine a 95% confidence interval for the meancost, μ, of all recent weddings in this country. The 95% confidence interval is from $ to $.b) Interpret your result in part (a). Choose the correct answer below. A. We can be 95% confident that any randomly selected recent wedding cost in this country is somewhere within the confidence interval. B.We can be 95% confident that the sample cost of recent weddings in this country is somewhere within the confidence interval. C. The mean cost, μ,of all recent weddings in this country is somewhere within the confidence interval. D.We can be 95% confident that the mean cost, μ,of all recent weddings in this country is somewhere within the confidence interval. c. Does the mean cost of all recent weddings in this country lie in the confidence interval obtained in part (a)? Explain your answer. A. Yes, since the confidence interval must contain the sample and population means. B. The population mean may or may not lie in this interval, but we can be 95% confident that it does. C. The sample mean may or may not lie in this interval, but we can be 95% confident that it does. D. No, since there is a 5% chance that the population mean does not lie in the confidence interval.

a) 95% confidence interval for the meancost, μ, of all recent weddings in this country = (22,550.95, 30,226.40)

.The 95% confidence interval is from $22,550.95 to $30,226.40.

b) For the interpretation of the result, option D is correct.

We can be 95% confident that the mean cost, μ, of all recent weddings in this country is somewhere within the confidence interval.

c) Option B is correct.

The population mean may or may not lie in this interval, but we can be 95% confident that it does.

Step-by-step explanation:

Sample size = 20

Sample Mean = $26,388.67

Sample Standard deviation = $8200

Confidence Interval for the population mean is basically an interval of range of values where the true population mean can be found with a certain level of confidence.

Mathematically,

Confidence Interval = (Sample mean) ± (Margin of error)

Sample Mean = 26,388.67

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 of the mean)

Critical value will be obtained using the t-distribution. This is because there is no information provided for the population mean and standard deviation.

To find the critical value from the t-tables, we first find the degree of freedom and the significance level.

Degree of freedom = df = n - 1 = 20 - 1 = 19.

Significance level for 95% confidence interval

(100% - 95%)/2 = 2.5% = 0.025

t (0.025, 19) = 2.086 (from the t-tables)

Standard error of the mean = σₓ = (σ/√n)

σ = standard deviation of the sample = 8200

n = sample size = 20

σₓ = (8200/√20) = 1833.6

99% Confidence Interval = (Sample mean) ± [(Critical value) × (standard Error of the mean)]

CI = 26,388.67 ± (2.093 × 1833.6)

CI = 26,388.67 ± 3,837.7248

99% CI = (22,550.9452, 30,226.3948)

99% Confidence interval = (22,550.95, 30,226.40)

a) 95% confidence interval for the meancost, μ, of all recent weddings in this country = (22,550.95, 30,226.40)

.The 95% confidence interval is from $22,550.95 to $30,226.40.

b) The interpretation of the confidence interval obtained, just as explained above is that we can be 95% confident that the mean cost, μ,of all recent weddings in this country is somewhere within the confidence interval

c) A further explanation would be that the population mean may or may not lie in this interval, but we can be 95% confident that it does.

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