Question

"A random survey of 927 adults in California found that 63% of them say they are likely to sleep when they stay home sick."

e. Construct a 95% confidence interval for p. (Be sure to follow the whole process).

f. Is it plausible to say that 70% of all California adults would sleep when they stay home sick?

g. Perform a hypothesis test to determine if more than 60% of adults would sleep when they stay home sick.

h. If we made an error here, then would it be a Type I or a Type II error?

Answer 1

e) At 95% confidence level, the critical value is z_0.025 = 1.96

The 95% confidence interval for population proportion is

+/- z_0.025 * sqrt((1 - )/n)

= 0.63 +/- 1.96 * sqrt(0.63 * (1 - 0.63)/927)

= 0.63 +/- 0.031

= 0.599, 0.661

f) No, it is not plausible to say that 70% of all California adults would sleep when they stay home sick , because the interval does not contain 0.70.

g) H₀: P < 0.60

    H₁: P > 0.60

The test statistic z = ( - P)/sqrt(P(1 - P)/n)

                              = (0.63 - 0.6)/sqrt(0.6 * 0.4/927)

                              = 1.86

P-value = P(Z > 1.86)

             = 1 - P(Z < 1.86)

             = 1 - 0.9686

             = 0.0314

Since the P-value is less than the significance level(0.0314 < 0.05), so we should reject the null hypothesis.

At 0.05 significance level, there is sufficient evidence to conclude that more than 60% of adults would sleep when they stay home sick.

h) Type I error.