The International Olympic Committee states that the female participation in the 2000 Summer Olympic Games was 42%, even with new sports such as weight lifting, hammer throw, and modern pentathlon being added to the Games. Broadcasting and clothing companies want to change their advertising and marketing strategies if the female participation increases at the next games. An independent sports expert arranged for a random sample of pre-Olympic exhibitions. The sports expert reported that 202 of 454 athletes in the random sample were women. Is this strong evidence that the participation rate may increase? Test an appropriate hypothesis and state your conclusion.
Let's set up the null and alternative hypothesis, find a formula to use, then go from there.
Null hypothesis:
Ho: p = .42 -->meaning: population proportion is equal to .42
Alternative hypothesis:
Ha: p > .42 -->meaning: population proportion is greater than .42
[ Note: The null hypothesis is what we suspect isn't true, while the alternative hypothesis is what we suspect is true (or claim to be true). The null hypothesis ALWAYS uses an equals sign. ]
Using a formula for a binomial proportion one-sample z-test with your data included, we have:
z = .44 - .42 -->test value (202/454 = .44) minus population value (.42)
divided by √[(.42)(.58)/454] -->.42 represents 42%, .58 represents 1-.42, and 454 is the sample size.
Finish the calculation and draw your conclusions. Remember if you reject the null and accept the alternative hypothesis, you will have determined there is enough evidence to support the claim that the participation rate has increased. If the null is not rejected, you cannot conclude a difference.
I hope this will help.
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To determine if there is strong evidence that the participation rate may increase, we can conduct a hypothesis test using the sample data provided.
Let's define the hypotheses for this test:
- Null hypothesis (H0): The female participation rate will remain the same as the 2000 Olympics (42%).
- Alternative hypothesis (Ha): The female participation rate will increase from the 2000 Olympics (greater than 42%).
We can use the chi-square test for independence to conduct this hypothesis test. Here are the steps:
Step 1: Set up the hypotheses:
H0: p = 0.42 (female participation rate remains the same)
Ha: p > 0.42 (female participation rate increases)
Step 2: Determine the significance level (α). Let's assume a significance level of α = 0.05.
Step 3: Gather the sample data:
The random sample consists of 202 women out of 454 athletes in the pre-Olympic exhibitions.
Step 4: Calculate the test statistic:
We can calculate the test statistic using the chi-square formula:
chi-square = (O - E)² / E
Where O is the observed frequency and E is the expected frequency.
Expected frequency for women = Total sample size * female participation rate
E = 454 * 0.42 = 190.68
chi-square = (202 - 190.68)² / 190.68 = 1.79
Step 5: Determine the critical value:
Since the alternative hypothesis is one-sided (p > 0.42), we will use the chi-square distribution table to find the critical value corresponding to α = 0.05 and degrees of freedom (df) = 1.
The critical value for α = 0.05 and df = 1 is approximately 3.841.
Step 6: Compare the test statistic to the critical value and make a decision:
In this case, the test statistic (1.79) is less than the critical value (3.841).
Step 7: Make a conclusion:
Since the test statistic (1.79) is less than the critical value (3.841), we fail to reject the null hypothesis (H0). There is not enough evidence to suggest that the female participation rate is significantly higher in the upcoming Olympic Games.
Therefore, based on the provided sample data, we cannot conclude that there is strong evidence the participation rate may increase.
To determine if there is strong evidence that the participation rate may increase, we need to conduct a hypothesis test.
First, let's set up the hypothesis:
Null hypothesis (H0): The female participation rate at the next Olympic Games is the same as the 2000 Summer Olympic Games.
Alternative hypothesis (H1): The female participation rate at the next Olympic Games is different from the 2000 Summer Olympic Games.
Next, we calculate the proportion of females in the random sample:
p̂ = number of females in the sample / sample size = 202 / 454 ≈ 0.445
To determine if this proportion is significantly different from 0.42 (the female participation rate in 2000), we can use a hypothesis test.
Assuming the null hypothesis is true, we can use a one-sample proportion z-test. The test statistic formula is:
z = (p̂ - p0) / sqrt(p0*(1-p0) / n)
Where:
p̂ is the observed proportion
p0 is the proportion under the null hypothesis
n is the sample size
Let's calculate the test statistic:
z = (0.445 - 0.42) / sqrt(0.42*(1-0.42) / 454) ≈ 0.686
Now, we need to determine the critical value for our chosen level of significance (e.g., α = 0.05).
Since this is a two-tailed test, we want to test whether the proportion is significantly different from either side. Therefore, we split the significance level in half, resulting in α/2 = 0.025.
We can use a standard normal distribution table, z-table, or software to find the critical value corresponding to the significance level of 0.025. For a two-tailed test at α = 0.05, the critical value is approximately ±1.96.
Now, compare the test statistic (0.686) with the critical value (±1.96).
If the test statistic falls within the critical region (outside the critical value range), we can reject the null hypothesis. If not, we fail to reject the null hypothesis.
Conclusion:
Since the test statistic (0.686) does not fall within the critical region of ±1.96, we fail to reject the null hypothesis. This means there is not strong evidence to suggest an increase in the female participation rate at the next Olympic Games compared to the 2000 Summer Olympic Games based on the given random sample.