Show all work. Any answer given with just an answer will be counted wrong. Also, Show the graph and shaded region as needed or it will be counted wrong. Make sure to answer the question using the steps of hypothesis testing. a.) state the hypothesis, b.) Select the level of significance, c.) Identify the test statistics, d.) Formulate a decision rule, e.) Work the problem arrive at a Decision, f.) State your decision, g.) What is your conclusion and how sure are you about your decision, use the p-value
1.You are about to fly to home for the holiday. Happy to get home, you tell your mother that if ticket prices have increased on average above $267 you won’t be able to come home. Looking online you collect the following sample below. Is it safe to conclude that the mean fare has increased? Better yet, will you be coming home for the holiday? Assume that N = 50 meaning that the mean and the STD for your sample of 13 would be exactly the same if you selected 50 observation.
Compare/contrast your conclusion by using the same calculation results to the other table, T or Z.
$321
$286
$290
$330
$310
$250
$270
$280
$299
$265
$291
$275
$281
2. Many of the grocery stores have used a self checkout system. You are thinking about firing one of the employees and buying one of the machines. You call Nicholas Hill in to the office and you tell him to count the number of people who use the self checkout for 15 days at a local store. If more than 100 people per day use the machine, Nicholas will be terminated. Below is a list of people who use the machine per day at the local store. Is it reasonable to fire Nicholas from his job at the .05 significance level? Compare/contrast your conclusion by using the same calculation results to the other table, T or Z.
120
108
120
114
118
91
118
92
104
104
112
97
118
108
117
1. a) State the hypothesis:
Null hypothesis (H0): The mean fare has not increased (μ ≤ $267)
Alternative hypothesis (Ha): The mean fare has increased (μ > $267)
b) Select the level of significance:
Level of significance (α) = 0.05
c) Identify the test statistic:
Since the sample size (n) is greater than 30, we can use the Z-test.
d) Formulate a decision rule:
Using a one-tailed test, we will reject the null hypothesis if the test statistic (Z-score) is greater than the critical value obtained from the Z-table for a significance level of 0.05.
e) Work the problem and arrive at a decision:
Step 1: Calculate the sample mean (x̄) and sample standard deviation (s) from the given data:
x̄ = (321 + 286 + 290 + 330 + 310 + 250 + 270 + 280 + 299 + 265 + 291 + 275 + 281) / 13 = 288.38
s = √[(∑(x - x̄)²) / (n - 1)] = √[(∑(x - 288.38)²) / (13 - 1)] = 22.46
Step 2: Calculate the test statistic (Z-score):
Z = (x̄ - μ) / (s / √n) = (288.38 - 267) / (22.46 / √13) = 3.03
Step 3: Look up the critical value for a one-tailed test at the 0.05 level of significance in the Z-table. Since it is not specified, we'll assume a right-tailed test. The critical value is 1.645.
Step 4: Compare the test statistic (Z-score) with the critical value:
Since 3.03 > 1.645, we reject the null hypothesis.
f) State your decision:
We reject the null hypothesis.
g) What is your conclusion and how sure are you about your decision, use the p-value:
Based on the sample data, we have sufficient evidence to conclude that the mean fare has increased. Our decision is based on a significance level of 0.05, meaning that we are 95% confident in our conclusion.
Comparing to the t-table:
Since the sample size is small (n < 30) and the population standard deviation is unknown, we can also perform a t-test. However, since the sample size is close to 50, we can still use the Z-test with minimal impact on the results.
2. a) State the hypothesis:
Null hypothesis (H0): The average number of people using the machine per day is ≤ 100
Alternative hypothesis (Ha): The average number of people using the machine per day is > 100
b) Select the level of significance:
Level of significance (α) = 0.05
c) Identify the test statistic:
Since the sample size (n) is greater than 30, we can use the Z-test.
d) Formulate a decision rule:
Using a one-tailed test, we will reject the null hypothesis if the test statistic (Z-score) is greater than the critical value obtained from the Z-table for a significance level of 0.05.
e) Work the problem and arrive at a decision:
Step 1: Calculate the sample mean (x̄) and sample standard deviation (s) from the given data:
x̄ = (120 + 108 + 120 + 114 + 118 + 91 + 118 + 92 + 104 + 104 + 112 + 97 + 118 + 108 + 117) / 15 = 109.67
s = √[(∑(x - x̄)²) / (n - 1)] = √[(∑(x - 109.67)²) / (15 - 1)] = 9.70
Step 2: Calculate the test statistic (Z-score):
Z = (x̄ - μ) / (s / √n) = (109.67 - 100) / (9.70 / √15) = 1.72
Step 3: Look up the critical value for a one-tailed test at the 0.05 level of significance in the Z-table. Since it is not specified, we'll assume a right-tailed test. The critical value is 1.645.
Step 4: Compare the test statistic (Z-score) with the critical value:
Since 1.72 > 1.645, we reject the null hypothesis.
f) State your decision:
We reject the null hypothesis.
g) What is your conclusion and how sure are you about your decision, use the p-value:
Based on the sample data, we have sufficient evidence to conclude that more than 100 people use the self-checkout machine per day. Our decision is based on a significance level of 0.05, meaning that we are 95% confident in our conclusion.
Comparing to the t-table:
Since the sample size is small (n < 30) and the population standard deviation is unknown, we can also perform a t-test. However, since the sample size is close to 30, we can still use the Z-test with minimal impact on the results.
1. To determine if the mean fare has increased, we will use the following steps of hypothesis testing:
a) State the hypothesis:
- Null Hypothesis (H0): The mean fare is not greater than $267.
- Alternative Hypothesis (Ha): The mean fare is greater than $267.
b) Select the level of significance:
Let's select a level of significance of α = 0.05.
c) Identify the test statistic:
Since we have a sample size (n) of 13 and the population standard deviation (σ) is unknown, we will use the t-test statistic.
d) Formulate a decision rule:
In this case, since it's a one-tailed test and we are looking for the mean fare to be greater than $267, our decision rule can be formulated as follows:
- If the t-test statistic is greater than the critical value from the t-distribution with (n-1) degrees of freedom, we reject the null hypothesis.
e) Work the problem and arrive at a decision:
Let's calculate the t-test statistic using the given sample data. Here are the data points:
$321, $286, $290, $330, $310, $250, $270, $280, $299, $265, $291, $275, $281
First, calculate the sample mean (x̄) and sample standard deviation (s) using the formula:
x̄ = (sum of all values) / n
s ≈ √[(sum of (x - x̄)²) / (n-1)]
After calculating, we find that x̄ ≈ $287.08 and s ≈ $24.95.
Next, calculate the t-test statistic using the formula:
t = (x̄ - μ) / (s / √n)
where μ is the hypothesized population mean.
Since the null hypothesis states that μ is not greater than $267, we can substitute $267 into the formula:
t = ($287.08 - $267) / ($24.95 / √13)
After calculating, we find that t ≈ 1.51.
Now, we need to find the critical value from the t-distribution with (n-1) degrees of freedom (n=13-1=12). Using a t-table or software, we find that the critical value for a one-tailed test at α = 0.05 is approximately 1.782.
f) State your decision:
Since the calculated t-test statistic (1.51) is not greater than the critical value (1.782), we fail to reject the null hypothesis.
g) What is your conclusion and how sure are you about your decision, using the p-value:
Since the p-value is not provided, we cannot determine its exact value. However, based on the t-test statistic calculated, the p-value would be larger than 0.05 (the chosen significance level) because the t-test statistic is not in the critical region. As a result, we fail to reject the null hypothesis, implying that it is safe to conclude that the mean fare has not increased. Therefore, you will be able to come home for the holiday.
We could also obtain similar conclusions by using the z-test statistic. However, in this case, since the population standard deviation is unknown, we need to use the t-test.
2. To determine if it is reasonable to fire Nicholas from his job, we will use the following steps of hypothesis testing:
a) State the hypothesis:
- Null Hypothesis (H0): The mean number of people who use the self checkout is not more than 100 per day.
- Alternative Hypothesis (Ha): The mean number of people who use the self checkout is more than 100 per day.
b) Select the level of significance:
Let's select a level of significance of α = 0.05.
c) Identify the test statistic:
Since we have a sample size (n) of 15 and the population standard deviation (σ) is unknown, we will use the t-test statistic.
d) Formulate a decision rule:
In this case, since it's a one-tailed test and we are looking for the mean number of people to be more than 100, our decision rule can be formulated as follows:
- If the t-test statistic is greater than the critical value from the t-distribution with (n-1) degrees of freedom, we reject the null hypothesis.
e) Work the problem and arrive at a decision:
Let's calculate the t-test statistic using the given sample data. Here are the daily counts:
120, 108, 120, 114, 118, 91, 118, 92, 104, 104, 112, 97, 118, 108, 117
First, calculate the sample mean (x̄) and sample standard deviation (s).
After calculating, we find that x̄ ≈ 110.27 and s ≈ 9.07.
Next, calculate the t-test statistic using the formula:
t = (x̄ - μ) / (s / √n)
where μ is the hypothesized population mean.
Since the null hypothesis states that μ is not more than 100, we can substitute 100 into the formula:
t = (110.27 - 100) / (9.07 / √15)
After calculating, we find that t ≈ 2.051.
Now, we need to find the critical value from the t-distribution with (n-1) degrees of freedom (n=15-1=14). Using a t-table or software, we find that the critical value for a one-tailed test at α = 0.05 is approximately 1.761.
f) State your decision:
Since the calculated t-test statistic (2.051) is greater than the critical value (1.761), we reject the null hypothesis.
g) What is your conclusion and how sure are you about your decision, using the p-value:
The p-value is not provided, so we cannot determine its exact value. However, based on the calculated t-test statistic, the p-value would be less than 0.05 (the chosen significance level) because the t-test statistic is in the critical region. As a result, we reject the null hypothesis, implying that it is reasonable to fire Nicholas from his job. Therefore, based on the given data, it is reasonable to terminate Nicholas from his job at the 0.05 significance level.
Similarly to the previous question, we could also obtain similar conclusions by using the z-test statistic. However, in this case, since the population standard deviation is unknown, we need to use the t-test.