A company realizes that the time a person holds in the line when calling their hot line is 4 minutes. As a way to improve its customer service, the firm implemented a new call procedure intended to reduce this time. After implementing it, the firm collected hold times from 75 randomly selected incoming calls to its hot line that were put on hold. These calls had sample mean hold time of 3.4 minutes. It is known that the standard deviation of the holding time is 2.3 minutes. State the null and alternative hypotheses the company would need to use and perform a valid statistical test to decide under the assumption of Normality with a significance level of \(\alpha = 0.05\).
Continuing with Problem 1, calculate the power of your test, plot it, and evaluate this function at \(\mu = 3\).
Write a function in R to solve both the previous problems. The function should:
The return of the function should include all the items in a list.
Take this vector as the sample, and assume that \(\sigma = 0.53\)
Test the hypothesis that \(\mu = 1.9\) versus the alternative that:
In testing that \(\mu = \mu_0\) vs. \(\mu < \mu_0\), assume that \(\sigma^2\) is unknown. What is the LR test in this case? What changes if the alternative is \(\mu > \mu_0\)? Implement these tests in R
Beginning accounting students need to learn to audit in a computerized environment. A sample of beginning accounting students took a test that is summarized by two scores shown in Table D.4computer_attitude_df`: the Computer Attitude Scale (CAS), based on 20 questions, and the Computer Anxiety Rating Scale (CARS), based on 19 questions. (Courtesy of Douglas Stein.) Males are coded as 1 and females as 0.
Gender <- c(0, 1, 0, 1, 1, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 0, 1,
1, 1, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1)
CAS <- c(2.85, 2.60, 2.20, 2.65, 2.60, 3.20, 3.65, 2.55, 3.15, 2.80, 2.40, 3.20, 3.05, 2.60, 3.35, 3.75, 3.00, 2.80,
3.30, 2.90, 2.60, 2.25, 1.90, 2.20, 2.30, 1.80, 3.05, 3.15, 2.80, 2.35, 3.70, 2.60, 3.50, 2.95, 2.80)
CARS <- c(2.90, 2.32, 1.00, 2.58, 2.58, 3.05, 3.74, 1.90, 3.32, 2.74, 2.37, 3.11, 3.32, 2.79, 2.95, 3.79, 3.26, 3.21,
3.47, 3.05, 2.68, 1.90, 1.84, 1.74, 2.58, 1.58, 2.47, 3.32, 2.90, 2.42, 3.47, 4.00, 3.42, 2.53, 2.68)
computer_attitude_df <- data.frame(
Gender = Gender,
CAS = CAS,
CARS = CARS
)Conduct an \(\alpha = .025\) level test of : \(\mu = 2.7\) versus \(\mu > 2.7\)
Let \((X_1, Y_1), \dots, (X_n, Y_n)\) be a random sample from a bivariate normal distribution with parameters \(\mu_X, \mu_Y, \sigma^2_X, \sigma^2_Y, \rho\). We are interested in testing
\[H_0: \mu_X = \mu_Y \text{ versus } H_1: \mu_X \neq \mu_Y .\]
Show that the random variables \(W_i = X_i - Y_i\) are iid \(N(\mu_W, \sigma^2_W)\).
Show that the above hypothesis can be tested with the statistic \[T_W = \frac{\bar{W}}{\sqrt{S^2_W / n}},\] where \(\bar{W} = \frac{1}{n}\sum_{i=1}^n W_i\) and \(S^2_W = \frac{1}{n-1}\sum_{i=1}^n (W_i - \bar{W})^2\). Furthermore, show that, under \(H_0\), \(T_W \sim t_{n-1}\)
This test is known as the paired-sample t test
Apply this test to the data in sleep. See ?sleep. Comment on your conclusions.