A coffee shop owner believes the shop’s average daily sales are $1,200, with a known population standard deviation of $50. A manager claims it’s higher because of their new initiatives & collects a sample of 25 days and finds an average daily sales of $1,230.(A)\left(A\right) Use α=0.10α=0.10 & the critical value method to test if the true mean daily sales are higher than $1200.H0:μ=H_0: μ= _____ Ha:μH_a: μ ______ x‾=\overline{x}= ______ σ=\sigma= ______ n=n= ______ z=z=Because test stat. (z)(z) is [ INSIDE | OUTSIDE ] rejection region, we [ REJECT | FAIL TO REJECT ] H0H_0. There is [ ENOUGH | NOT ENOUGH ] evidence to conclude that…

Because test stat. ( z ) (z) ( z ) is INSIDE rejection region, we REJECT H 0 H_0 H 0 . There is ENOUGH evidence to conclude that μ > 1200.

A coffee shop owner believes the shop’s average daily sales are $1,200, with a known population standard deviation of $50. A manager claims it’s higher because of their new initiatives & collects a sample of 25 days and finds an average daily sales of $1,230. ( A ) \left(A\right) Use α = 0.10 α=0.10 & the critical value method to test if the true mean daily sales are higher than $1200. H 0 : μ = H_0: μ= _____ H a : μ H_a: μ ______ x ‾ = \overline{x}= ______ σ = \sigma= ______ n = n= ______ z = z= Because test stat. ( z ) (z) is [ INSIDE | OUTSIDE ] rejection region, we [ REJECT | FAIL TO REJECT ] H 0 H_0 . There is [ ENOUGH | NOT ENOUGH ] evidence to conclude that…

Welcome to the practice. In this problem, we have a coffee shop owner believing the shop's average daily sales are $1,200 with a known population standard deviation of $50. A manager claims it's higher because of their new initiatives and collects a sample of twenty five days and finds an average daily sales of $1,230. In part a, we have that alpha is point one, and we'd like to use the critical value method to test if the true mean daily sales are higher than $1,200. Now this is a hypothesis test, so we're gonna begin by writing our hypotheses, where the null hypothesis is that mu, the population mean, or average daily sales, is what we expect it to be, and that expectation is $1,200. So our null hypothesis is mu equals 1,200. The alternative hypothesis is what we're trying to test or find evidence for, and we can see that we're trying to see if the true mean daily sales are higher than 1,200. So our alternative hypothesis is that mu is greater than 1,200. Now that we have our hypothesis, it's time to find our test statistic, which will be a z score because we know the value of sigma, the population standard deviation. But before we just find that z score, it is gonna be a good idea to get some key values from the problem so that they're easy to plug in. We'll need x bar, our sample mean, which is 1,230. We'll need sigma, the population standard deviation, which is 50, and we need n, the sample size, which is 25. Now that we have those values, we can plug them into our formula to get our z score. We get twelve thirty minus the expected population mean, 1,200, divided by 50 divided by the square root of 25. And you can definitely use a calculator to help you simplify this, you will end up getting three as your z score. Now that we have our test statistic, it's time to find our critical value. Looking at our alternative hypothesis, we can see that we have a greater than symbol, which means that we're doing a right tailed hypothesis test. That means that we're gonna have one critical value, and it is going to be the z value that has alpha, which is 0.1, as its right tail probability or area. You can use either a z table or technology to get your z critical value. It will end up being about 1.28. So let's turn our attention to our graphic. We can see that our critical value of 1.28 is gonna fall about here, 1.28. And because we're doing a right tailed hypothesis test, the rejection region is going to be the area to the right of our critical value, the right tail of that 1.28. Now, we want to see where our test statistic of three would fall and if it falls in the rejection region. While three is greater than 1.28, it falls about here, and we can see that it does fall in that right tail, so it is in the rejection region. So now it's time for our conclusion. Because the test statistic, our z score, is inside the rejection region, we reject our null hypothesis, so there is enough evidence to conclude that the alternative hypothesis is true and that mu is greater than 1,200. Or, in other words, there is enough evidence to conclude that the average daily sales is greater than $1,200. Great work.

A coffee shop owner believes the shop’s average daily sales are $1,200, with a known population standard deviation of $50. A manager claims it’s higher because of their new initiatives & collects a sample of 25 days and finds an average daily sales of $1,230. ( A ) \left(A\right) Use α = 0.10 α=0.10 & the critical value method to test if the true mean daily sales are higher than $1200. H 0 : μ = H_0: μ= _____ H a : μ H_a: μ ______ x ‾ = \overline{x}= ______ σ = \sigma= ______ n = n= ______ z = z= Because test stat. ( z ) (z) is [ INSIDE | OUTSIDE ] rejection region, we [ REJECT | FAIL TO REJECT ] H 0 H_0 . There is [ ENOUGH | NOT ENOUGH ] evidence to conclude that…

Because test stat. ( z ) (z) ( z ) is INSIDE rejection region, we REJECT H 0 H_0 H 0 . There is ENOUGH evidence to conclude that μ > 1200.

Because test stat. ( z ) (z) ( z ) is INSIDE rejection region, we REJECT H 0 H_0 H 0 . There is ENOUGH evidence to conclude that μ = 1200.

Because test stat. ( z ) (z) ( z ) is OUTSIDE rejection region, we FAIL TO REJECT H 0 H_0 H 0 . There is NOT ENOUGH evidence to conclude that μ = 1200.

Because test stat. ( z ) (z) ( z ) is OUTSIDE rejection region, we FAIL TO REJECT H 0 H_0 H 0 . There is NOT ENOUGH evidence to conclude that μ > 1200.

9. Hypothesis Testing for One Sample

Critical Values and Rejection Regions

Statistics for Business

9. Hypothesis Testing for One Sample

Critical Values and Rejection Regions

Struggling with Statistics for Business?

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Multiple Choice

A coffee shop owner believes the shop’s average daily sales are \$1,200, with a known population standard deviation of \$50. A manager claims it’s higher because of their new initiatives & collects a sample of 25 days and finds an average daily sales of \$1,230.
$(A)$ Use $α = 0.10$ & the critical value method to test if the true mean daily sales are higher than \$1200.
$H_{0} : μ =$ _____ $H_{a} : μ$ $\overline{x} =$ $σ =$ $n =$ $z =$
Because test stat. $(z)$ is [ INSIDE | OUTSIDE ] rejection region, we [ REJECT | FAIL TO REJECT ] $H_{0}$ . There is [ ENOUGH | NOT ENOUGH ] evidence to conclude that…

Because test stat. $(z)$ is INSIDE rejection region, we REJECT $H_0$ . There is ENOUGH evidence to conclude that μ = 1200.

Because test stat. $(z)$ is INSIDE rejection region, we REJECT $H_0$ . There is ENOUGH evidence to conclude that μ > 1200.

Because test stat. $(z)$ is OUTSIDE rejection region, we FAIL TO REJECT $H_0$ . There is NOT ENOUGH evidence to conclude that μ = 1200.

Because test stat. $(z)$ is OUTSIDE rejection region, we FAIL TO REJECT $H_0$ . There is NOT ENOUGH evidence to conclude that μ > 1200.

Verified step by step guidance

Identify the null hypothesis ($H_0$) and the alternative hypothesis ($H_a$) based on the context of the problem. Typically, $H_0$ represents the status quo or no effect, and $H_a$ represents the claim you want to test.

Determine the significance level ($\alpha$), which is the probability of rejecting the null hypothesis when it is actually true. Common values are 0.05, 0.01, or 0.10.

Find the critical value(s) corresponding to the chosen significance level and the type of test (left-tailed, right-tailed, or two-tailed). Use the standard normal distribution ($Z$) or $t$-distribution table depending on whether the population standard deviation is known and the sample size.

Calculate the test statistic using the sample data. For example, if using a $Z$-test, the formula is: \[\text{Test Statistic} = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}}\] where $\bar{x}$ is the sample mean, $\mu_0$ is the hypothesized population mean, $\sigma$ is the population standard deviation, and $n$ is the sample size.

Compare the test statistic to the critical value(s) to decide whether to reject or fail to reject the null hypothesis. If the test statistic falls into the critical region, reject $H_0$; otherwise, do not reject $H_0$.

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