An appliance manufacturer stockplies washers and dryers in a large warehouse for shipment to retail stores. Sometimes in handling then the appliances get damaged. Even though the damage may be minor, the company must sell those machines at drastically reduced prices. The company goal is to keep the level of damaged machines below 2%. One day an inspector randomly checks 60 washers and finds that 5 of them have scratches or dents. Is this strong evidence that the warehouse is failing to meet the company goal? Test an appropriate hypothesis and state you conclusion. Be sure the appropriate assumptions and conditions are satisfied before you proceed.
p= proportion of washes that are damaged
H0:p=.002 Ha:p>.002
Plausible Independence: assume washes are independent of each other
success/failure (.002)(.60)=12>10
10% condition: there are at least 600 washers
z=.0834-.05 (divided by)/square root (.002)(.098) (divided by) 60
equals=.00122
okay now that is how much I have been able to figure out but I don't know if all of that is correct and I need to figure out the p-value and a conclusion. please help?
Ok, I'm a little lost in your answer. I would say:
H0:p=.02, Ha:p>.02
The expected mean number of damages is p*n = .02*60 = 1.2
The formula for the standard deviation for a binominal distribution is sqrt(n*p*q) (where q=1-p)
so, sd = sqrt(60*.02*.98) = 1.0844
the alpha for 95% confidence interval (one tailed) is 1.645. We will reject the null hypothesis if we number of damaged washers is greater than 1.2+1.645*sd = 2.98. Since we observed 5 damage washer, reject the null hypothesis.
I hope this helps.
To determine if the warehouse is failing to meet the company goal, we can conduct a hypothesis test based on the given information.
First, let's define the hypotheses:
- Null Hypothesis (H0): The proportion of damaged machines in the warehouse is 0.002 (p = 0.002).
- Alternative Hypothesis (Ha): The proportion of damaged machines in the warehouse is greater than 0.002 (p > 0.002).
Next, we need to check if the assumptions and conditions for conducting a hypothesis test are satisfied:
1. Plausible Independence: Assuming that the washers are independent of each other, you mentioned this assumption, which is reasonable.
2. Success/Failure: To use the normal distribution approximation, we need a sufficiently large sample. In this case, we have 60 washers (n = 60), and the expected number of successes is equal to the hypothesized proportion multiplied by the sample size (0.002 * 60 = 0.12). Since this is greater than 10, the success/failure condition is satisfied.
3. Random Sampling: The problem states that the inspector checks the washers randomly, so this condition is satisfied.
Now, we can calculate the test statistic (z-score) and the p-value:
The test statistic, z, is calculated using the formula:
z = (p̂ - p0) / √(p0 * (1 - p0) / n)
Given that 5 out of the 60 washers inspected were damaged, p̂ (the sample proportion) is 5/60 = 0.0833. Plugging in the values, we get:
z = (0.0833 - 0.002) / √(0.002 * (1 - 0.002) / 60)
z ≈ 1.22
To find the p-value (the probability of observing a test statistic as extreme or more extreme than the one obtained, assuming the null hypothesis is true), we need to calculate the area under the standard normal curve to the right of z = 1.22. Using a z-table or calculator, we find that the p-value is approximately 0.111.
Finally, we can make a conclusion:
Since the p-value (0.111) is greater than the commonly chosen significance level of 0.05, we fail to reject the null hypothesis. This means that the evidence is not strong enough to conclude that the warehouse is failing to meet the company goal of keeping the level of damaged machines below 2%.
To test the hypothesis, we need to calculate the p-value and make a conclusion based on the significance level.
To calculate the p-value, we can use the standard normal distribution.
The observed proportion of damaged washers is 5/60 = 0.0833.
The null hypothesis states that the proportion of damaged washers is 0.002.
Using the formula for a one-sample z-test, the test statistic is calculated as follows:
z = (0.0833 - 0.002) / sqrt((0.002 * 0.998) / 60)
Computing this, we find a test statistic of approximately 67.96.
Since the alternative hypothesis is one-sided (Ha: p > 0.002), we need to find the area to the right of the test statistic under the standard normal curve. This area represents the p-value.
The p-value can be calculated using a standard normal distribution table or a calculator. The p-value is the probability of observing a test statistic as extreme as 67.96 or greater. In this case, the p-value would be extremely close to 0.
Since the p-value is less than the significance level (usually 0.05), we reject the null hypothesis. This means there is strong evidence to suggest that the warehouse is failing to meet the company goal of keeping the level of damaged machines below 2%.
In conclusion, based on the data and the calculated p-value, we can conclude that the evidence strongly suggests that the warehouse is not meeting the company goal.