The problem involves hypothesis testing for three different scenarios. Scenario 1: An e-commerce business wants to test if the proportion of satisfied customers is significantly less than 80% after launching a new website design. Scenario 2: A pharmaceutical company wants to verify if the mean Vitamin C content in tablets remains at 500 mg. The standard deviation is known to be 5 mg and a sample of 50 tablets is taken. Scenario 3: A coffee shop chain wants to assess if the average duration customers spend in their stores is less than 45 minutes. A sample of 20 customers was taken, but the standard deviation is unknown.

1. Problem Description

The problem involves hypothesis testing for three different scenarios.

Scenario 1: An e-commerce business wants to test if the proportion of satisfied customers is significantly less than 80% after launching a new website design.

Scenario 2: A pharmaceutical company wants to verify if the mean Vitamin C content in tablets remains at 500 mg. The standard deviation is known to be 5 mg and a sample of 50 tablets is taken.

Scenario 3: A coffee shop chain wants to assess if the average duration customers spend in their stores is less than 45 minutes. A sample of 20 customers was taken, but the standard deviation is unknown.

2. Solution Steps

Due to the prompt's request not to use bold symbols, I will use underlines instead to represent underscores if necessary. I will provide general steps for each scenario, leaving the specifics to be filled in with hypothetical values or actual data.

Scenario 1: E-commerce Customer Satisfaction Proportion

1. Take a sample: Let's assume a sample size of $n=100$ customers. Let's say $x=70$ customers are satisfied. So, the sample proportion is $\hat{p} = \frac{x}{n} = \frac{70}{100} = 0.7$.

2. State the null and alternative hypotheses:

* Null hypothesis ($H_0$): $p \ge 0.8$ (The proportion of satisfied customers is greater than or equal to 80%)

* Alternative hypothesis ($H_1$): $p < 0.8$ (The proportion of satisfied customers is less than 80%)

3. Select the level of significance: Let's choose $\alpha = 0.05$.

4. Select the appropriate test statistic: Since we are testing a single population proportion and $n$ is reasonably large, we can use the z-test statistic:

$z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}$

where $\hat{p}$ is the sample proportion, $p_0$ is the hypothesized proportion (0.8), and $n$ is the sample size.

5. Calculate the value of the test statistic:

$z = \frac{0.7 - 0.8}{\sqrt{\frac{0.8(1-0.8)}{100}}} = \frac{-0.1}{\sqrt{\frac{0.16}{100}}} = \frac{-0.1}{0.04} = -2.5$

6. Formulate the decision rule:

* Critical value method: For a left-tailed test with $\alpha = 0.05$, the critical value is $z_{\alpha} = -1.645$. We reject $H_0$ if $z < -1.645$.

* P-value method: Calculate the p-value, which is $P(Z < -2.5) = 0.0062$ (using a standard normal table or calculator). We reject $H_0$ if the p-value is less than $\alpha$.

7. Decide based on your decision rule:

* Critical value method: Since $-2.5 < -1.645$, we reject $H_0$.

* P-value method: Since $0.0062 < 0.05$, we reject $H_0$.

8. Interpret the decision: There is sufficient evidence to conclude that the proportion of satisfied customers is significantly less than 80%.

Scenario 2: Vitamin C Content in Tablets

1. Take a sample mean: Let's assume the sample mean Vitamin C content is $\bar{x} = 498$ mg. The population standard deviation is $\sigma = 5$ mg, and the sample size is $n = 50$.

2. State the null and alternative hypotheses:

* Null hypothesis ($H_0$): $\mu = 500$ (The mean Vitamin C content is 500 mg)

* Alternative hypothesis ($H_1$): $\mu \ne 500$ (The mean Vitamin C content is not 500 mg)

3. Select the level of significance: Let's choose $\alpha = 0.05$.

4. Select the appropriate test statistic: Since the population standard deviation is known, we can use the z-test:

$z = \frac{\bar{x} - \mu_0}{\frac{\sigma}{\sqrt{n}}}$

where $\bar{x}$ is the sample mean, $\mu_0$ is the hypothesized population mean (500), $\sigma$ is the population standard deviation, and $n$ is the sample size.

5. Calculate the value of the test statistic:

$z = \frac{498 - 500}{\frac{5}{\sqrt{50}}} = \frac{-2}{\frac{5}{\sqrt{50}}} = \frac{-2}{0.707} = -2.828$

6. Formulate the decision rule:

* Critical value method: For a two-tailed test with $\alpha = 0.05$, the critical values are $z_{\alpha/2} = -1.96$ and $z_{1-\alpha/2} = 1.96$. We reject $H_0$ if $z < -1.96$ or $z > 1.96$.

* P-value method: Calculate the p-value, which is $2 * P(Z < -2.828) = 2 * 0.0023 = 0.0046$ (using a standard normal table or calculator). We reject $H_0$ if the p-value is less than $\alpha$.

7. Decide based on your decision rule:

* Critical value method: Since $-2.828 < -1.96$, we reject $H_0$.

* P-value method: Since $0.0046 < 0.05$, we reject $H_0$.

8. Interpret the decision: There is sufficient evidence to conclude that the mean Vitamin C content is significantly different from 500 mg.

Scenario 3: Coffee Shop Customer Duration

1. Sample information: The sample size is $n = 20$. Let's assume the sample mean duration is $\bar{x} = 42$ minutes, and the sample standard deviation is $s = 8$ minutes.

2. State the null and alternative hypotheses:

* Null hypothesis ($H_0$): $\mu \ge 45$ (The average duration is greater than or equal to 45 minutes)

* Alternative hypothesis ($H_1$): $\mu < 45$ (The average duration is less than 45 minutes)

3. Select the level of significance: Let's choose $\alpha = 0.05$.

4. Select the appropriate test statistic: Since the population standard deviation is unknown and the sample size is small, we use the t-test:

$t = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}}$

where $\bar{x}$ is the sample mean, $\mu_0$ is the hypothesized population mean (45), $s$ is the sample standard deviation, and $n$ is the sample size. The degrees of freedom are $df = n-1 = 20-1 = 19$.

5. Calculate the value of the test statistic:

$t = \frac{42 - 45}{\frac{8}{\sqrt{20}}} = \frac{-3}{\frac{8}{4.472}} = \frac{-3}{1.789} = -1.677$

6. Formulate the decision rule:

* Critical value method: For a left-tailed t-test with $\alpha = 0.05$ and $df=19$, the critical value is $t_{\alpha, df} = -1.729$. We reject $H_0$ if $t < -1.729$.

* P-value method: Calculate the p-value, which is $P(T < -1.677)$ with $df = 19$. Using a t-table or calculator, the p-value is approximately $0.054$. We reject $H_0$ if the p-value is less than $\alpha$.

7. Decide based on your decision rule:

* Critical value method: Since $-1.677 > -1.729$, we fail to reject $H_0$.

* P-value method: Since $0.054 > 0.05$, we fail to reject $H_0$.

8. Interpret the decision: There is not enough evidence to conclude that the average duration customers spend in the coffee shop is less than 45 minutes.

9. Final Answer

Scenario 1: There is sufficient evidence to conclude that the proportion of satisfied customers is significantly less than 80%.

Scenario 2: There is sufficient evidence to conclude that the mean Vitamin C content is significantly different from 500 mg.

Scenario 3: There is not enough evidence to conclude that the average duration customers spend in the coffee shop is less than 45 minutes.