Please Give Me A Solution For The Questions.

Please Give Me A Solution For The Questions1 Suppose 60 Of People

Suppose 60% of people now own smartphones. Assume the Central Limit Theorem conditions are satisfied. In any given sample of any size, what percentage of people would we expect to own smartphones?

There is a large bag of colored marbles. 20% are green and 80% are red, draw a marble then put it back so the next draw is independent of the previous. What is the probability that 5 of the next 10 are green?

The rate of asthma attacks in children ages 5-14 was 0.074 in 2009. A researcher is curious if advances in technology have reduced this proportion in the half-decade since and decides to perform a hypothesis test. Choose the correct null and alternative hypotheses:

  1. Null Hypothesis p=0.0744; Alternative Hypothesis p>0.0744
  2. Null Hypothesis p=0.0744; Alternative Hypothesis p
  3. Null Hypothesis p>0.0744; Alternative Hypothesis p

In the researcher’s sample of 1,500 children, the sample proportion of children was 0.0721. Compute the z-statistic of this sample proportion. What is the p-value?

Paper For Above instruction

The following analysis addresses the provided statistical questions concerning proportions, binomial probabilities, and hypothesis testing in healthcare and behavioral contexts. Each problem will be tackled systematically—from expected values under the Central Limit Theorem to probability calculations and hypothesis testing procedures, including the calculation of z-statistics and p-values.

1. Expected Percentage of Smartphone Ownership

Given that 60% of the population owns smartphones, and the Central Limit Theorem (CLT) conditions are satisfied, the expected proportion of people owning smartphones in any given sample remains 60%. This is because, under the CLT, for sufficiently large samples, the sampling distribution of the sample proportion is approximately normally distributed with a mean equal to the population proportion (p=0.60). Consequently, regardless of the sample size, the expected value (mean) of the sample proportion of smartphone owners is 60%. Therefore, we anticipate that about 60% of any sample would own smartphones, aligning with the population proportion.

2. Probability of Green Marbles in a Binomial Setting

Considering a large bag with 20% green marbles and 80% red marbles, and assuming each draw is independent with replacement, the probability of drawing a green marble in a single draw is 0.20. To find the probability of getting exactly 5 green marbles in 10 draws, we model this using the binomial distribution with parameters n=10 and p=0.20.

The probability mass function (PMF) of the binomial distribution gives:

P(X=5) = C(10,5) (0.20)^5 (0.80)^5

where C(10,5) is the binomial coefficient (10 choose 5), which equals 252.

Calculating this value:

P(X=5) ≈ 252 (0.20)^5 (0.80)^5 ≈ 252 0.00032 0.32768 ≈ 0.0264, or approximately 2.64%.

3. Hypotheses Testing: Asthma Attack Rates

The problem states that the rate of asthma attacks in children aged 5-14 was 0.074 in 2009. The researcher hypothesizes that technological advancements have reduced this rate. The hypotheses are formulated as:

  1. Option b): Null hypothesis p=0.0744; Alternative hypothesis p

This reflects a test for a reduction in the proportion, where the null posits the proportion remains at 0.0744 (noting the slight difference in decimal precision from 0.074), and the alternative suggests a decrease (p

4. Computing the Z-Statistic

The sample proportion is p̂ = 0.0721, from n=1,500 children. The standard error (SE) of the proportion under the null hypothesis is calculated as:

SE = √[p₀(1 - p₀)/n] = √[0.0744 (1 - 0.0744)/1500] ≈ √[0.0744 0.9256 / 1500] ≈ √[0.0689 / 1500] ≈ √0.0000459 ≈ 0.00678

The z-statistic is given by:

z = (p̂ - p₀) / SE = (0.0721 - 0.0744) / 0.00678 ≈ -0.0023 / 0.00678 ≈ -0.339

This negative z value indicates the sample proportion is slightly less than the hypothesized proportion, though close.

5. Calculating the P-Value

Since the alternative hypothesis is p

Conclusion

In summary, the expected proportion of smartphone owners in any sample is 60%, the probability of exactly 5 green marbles in 10 draws is approximately 2.64%, the hypotheses for reduced asthma attack rates are correctly framed as a one-tailed test with p

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