Of 150 patients randomly selected by a hospital, 90 of them said that their health insurance was not adequate.
(a) Construct a 99% confidence interval for the proportion of patients who would say that their health insurance was not adequate.
(b) Can we say that more than 50% of all patients of the hospital did not have adequate health insurance? Test at the 1% significance level.
a)
Standard error of the mean = SEM = √x(N-x)/N3 = 0.040
α = (1-CL)/2 = 0.005
Standard normal deviate for α = Zα = 2.576
Proportion of positive results = P = x/N = 0.600
Lower bound = P - (Zα*SEM) = 0.497
Upper bound = P + (Zα*SEM) = 0.703
b) No, because interval contains values less than 50% also
Of 150 patients randomly selected by a hospital, 90 of them said that their health insurance...
11: In a Gallup poll of 1000 randomly selected adults, 347 of them said they were underpaid. Construct a 95% confidence interval estimate of the percentage (proportion) of all adults who say that they are underpaid. Round to the nearest hundredth. 12: If you are using a ? −test when testing a claim about a population mean and the sample size is 53, the number of degrees of freedom is ________.
Of 92 adults selected randomly from one town, 61 have health insurance. Find a 90% confidence interval for the true proportion of all adults in the town who have health insurance. 0.566 < p < 0.760 0.548 < p < 0.778 0.536 < p < 0.790 0.582 < p < 0.744
In a Gallup poll, 1025 randomly selected adults were surveyed and 29% of them said that they used the Internet for shopping at least a few times a year. Find a 99% confidence interval estimate of the proportion of adults who use the Internet for shopping. (5 points)
1-Prop Conf. Int. 07 or 94 adults selected randomly from one town, 67 have health insurance. Find a 90% confidence interval for the true proportion of all adults in the town who have health insurance. Use Interval Notation with decimal rounded to the thousandths, Answer
In a Gallup poll, 1025 randomly selected adults were surveyed and 298 of them said that they used the internet for shopping at least a few times a year. a) Find the point estimate of the percentage of adults who use the internet for shopping. b) Find a 99% confidence interval estimate of the percentage of adults who use the internet for shopping.
Confidence Intervals: A group of 50 randomly selected JWU students have a mean age of 20.5 years. Assume the population standard deviation is 1.5 years. Construct a 99% confidence interval for the JWU population mean age. State your answer. (Zc 2.57) 1. 2. Construct a 90% confidence interval for the population mean, . Assume the population has a 2 normal distribution. A random sample of 20 JWU college students has mean annual earnings of 0 $3310 with a standard deviation...
You conduct a poll of 500 randomly selected city residents, asking them if they own an automobile, 280 say they do own an automobile. Construct a 94% confidence interval for the population proportion of city residents who own an automobile. Options- 5 to 6 .519 to .603 .518 to .602
of 146 adults selected randomly from one town, 25 of them smoke. Construct a 99% confidence interval for the true percentage of all adults in the town that smoke. Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p. Points: 5 5) Of 146 adults selected randomly from one town, 25 of them smoke. Construct a 99% confidence interval for the true percentage of all adults in the town that smoke....
In a study conducted in March 2020, 144 of 200 randomly selected students said that Liverpool football team were their favourite team. (a) Construct a 98% confidence interval to estimate the true proportion of students who said that Liverpool football team were their favourite team. (b) How large a sample would be necessary to estimate the true proportion of with a margin of error of 5%, with 95% confidence?
Suppose a hospital would like to estimate the proportion of patients who feel that physicians who care for them always communicated effectively when discussing their medical care. A pilot sample of 40 patients found that 22 reported that their physician communicated effectively. Determine the additional number of patients that need to be sampled to construct a 99% confidence interval with a margin of error equal to 8% to estimate this proportion. the additional patients that need to be sampled is...