Question

The mean survival time for patients with a particular form of disease is 4 years. A...

The mean survival time for patients with a particular form of disease is 4 years. A new treatment is tried on 20 patients with this particular disease and their average survival time is 4.7 years with a standard deviation of 1.5 years.

(a) State the direction of the alternative hypothesis to test if the new treatment constitutes a significant improvement in survival time. Type gt (greater than), ge (greater than or equal to), lt (less than), le (less than or equal to) or ne (not equal to) as appropriate in the box .

(b) State the critical value to conduct the test at 5% level of significance (three decimal places). Type only the numeric value in the box . (No need to specify whether it is tcrit or Zcrit)

(c) What is the calculated test statistic (three decimal places)? Type only the numeric value in the box . (No need to specify whether it is tcalc or Zcalc.)

(d) If the sampling error for a 90% confidence interval was required to be no more than 9 months, how large a sample would be necessary? (Assume 1.5 years is a good estimate of population standard deviation.) Type only the numeric value in the box .

Homework Answers

Answer #2

Solution:-

(a) Here the direction is on positive side .

Answer is greater than.

Years

years

(b)

Here sample size is  n = 20

Degree of freedom =

df= n-1 = 20 - 1

df=19

critical value

(c)

Here standard error

Stabdard Error=.3354

(d)

Margin of error =9 months = 0.75 year

sample size = n

critical value = NORMSINV(0.50 + 0.90/2) = 1.645

n = 10.82 or n = 11

answered by: anonymous
Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
A taxi company wants to predict travel time to the airport (in minutes) based on the...
A taxi company wants to predict travel time to the airport (in minutes) based on the distance from various locations in the city (in km). The data used involved distances ranging from 4km to 51km. A simple linear regression gave the following results. Regression Statistics R Square 0.917 Standard Error 2.579 Observations 12 Coefficients Standard Error Intercept 3.37 2.61 Distance 1.46 0.14 (a) What is the sample size used in this regression analysis? Type only the numerical value in the...
Macarthurs, a manufacturer of ropes used in abseiling, wished to determine if the production of their...
Macarthurs, a manufacturer of ropes used in abseiling, wished to determine if the production of their ropes was performing according to their specifications. All ropes being manufactured were required to have an average breaking strength of 228.5 kilograms and a standard deviation of 27.3 kilograms. They planned to test the breaking strength of their ropes using a random sample of forty ropes and were prepared to accept a Type I error probability of 0.01. 1. State the direction of the...
The Australian Medical Association believed that the Health Minister's recent statement claiming that 75% of doctors...
The Australian Medical Association believed that the Health Minister's recent statement claiming that 75% of doctors supported the reforms to Medicare was incorrect. It thought that the actual support for the reforms was lower than this. The Association's President suggested the best way to test this was to survey 150 members, selected through a random sample, on the issue. She indicated that the Association would be prepared to accept a Type I error probability of 0.05. 1. State the direction...
An environmental economist wished to determine the market demand function for tradable permits in carbon dioxide...
An environmental economist wished to determine the market demand function for tradable permits in carbon dioxide emissions from experimental data she had collected. Using 13 different values for the price per tonne in dollars (P), the following was estimated: Q = 6022.5 - 5.833 P Standard error of the estimate: 378.42 Sum of squares of price: 42681.0 where Q is the number of thousand tonnes of carbon dioxide demanded at the price P. 1. State the direction of the alternative...
The manager of Tile Central believes that the Virginia store is more profitable, in terms of...
The manager of Tile Central believes that the Virginia store is more profitable, in terms of weekly profit per full-time employee, than the Clayfield store. Random samples of size 17 and 16 accounting records for Clayfield and Virginia, respectively, were taken. These yielded average weekly profit per full-time employee for Clayfield and Virginia of $278.80 and $304.68, respectively, and variances of 881.32 and 993.45 dollars squared. It is assumed that the population variances are equal and the distribution of weekly...
The manager of Tile Central believes that the Virginia store is more profitable, in terms of...
The manager of Tile Central believes that the Virginia store is more profitable, in terms of weekly profit per full-time employee, than the store at Clayfield. Random samples of 13 and 15 accounting records for Clayfield and Virginia, respectively, were taken. These yielded average weekly profit per full-time employees for Clayfield and Virginia of $288.13 and $310.05, respectively, and variances of 889.52 $ squared and 1057.28 $ squared. It is assumed that the population variances are the same and the...
A recent study by the World Bank wished to determine whether there was a relationship between...
A recent study by the World Bank wished to determine whether there was a relationship between the abundance of natural resources in a country and its long term rate of economic growth. Their study used 40 (n) countries over a long period to 1995. Letting Y be the rate of growth measured as a percentage and X be a measure of natural resource abundance, the following relationship between the two variables was estimated. Standard errors of b0 and b1 are...
An observational study of Alzheimer's disease (AD) obtained data from 12 AD patients exhibiting moderate dementia...
An observational study of Alzheimer's disease (AD) obtained data from 12 AD patients exhibiting moderate dementia and selected a group of 11 control individuals without AD. AD is a progressive neurodegenerative disease of the elderly and advancing age is known to be a primary risk factor in AD diagnosis. Therefore, it was crucial for the study's credibility to examine whether the ages in the AD group might be significantly different than in the control group. The ages of the subjects...
An observational study of Alzheimer's disease (AD) obtained data from 8 AD patients exhibiting moderate dementia...
An observational study of Alzheimer's disease (AD) obtained data from 8 AD patients exhibiting moderate dementia and selected a group of 6 control individuals without AD. AD is a progressive neurodegenerative disease of the elderly and advancing age is known to be a primary risk factor in AD diagnosis. Therefore, it was crucial for the study's credibility to examine whether the ages in the AD group might be significantly different than in the control group. The ages of the subjects...
An observational study of Alzheimer's disease (AD) obtained data from 10 AD patients exhibiting moderate dementia...
An observational study of Alzheimer's disease (AD) obtained data from 10 AD patients exhibiting moderate dementia and selected a group of 7 control individuals without AD. AD is a progressive neurodegenerative disease of the elderly and advancing age is known to be a primary risk factor in AD diagnosis. Therefore, it was crucial for the study's credibility to examine whether the ages in the AD group might be significantly different than in the control group. The ages of the subjects...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT