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Past experience has shown that the heights of a certain variety of rose bush have been normally distributed with mean 85.0 cm. A new fertiliser is used and it is hoped that this will increase the heights. In order to test whether this is the case, a botanist records the heights, x cm, of a large random sample of n rose bushes and calculates that mu = 85.7 and s = 4.8, where mu is the sample mean and s^2 is an unbiased estimate of the population variance. The botanist then carries out an appropriate hypothesis test. The value of n = 150.
 one year ago
 one year ago
Past experience has shown that the heights of a certain variety of rose bush have been normally distributed with mean 85.0 cm. A new fertiliser is used and it is hoped that this will increase the heights. In order to test whether this is the case, a botanist records the heights, x cm, of a large random sample of n rose bushes and calculates that mu = 85.7 and s = 4.8, where mu is the sample mean and s^2 is an unbiased estimate of the population variance. The botanist then carries out an appropriate hypothesis test. The value of n = 150.
 one year ago
 one year ago

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orderBest ResponseYou've already chosen the best response.0
Using this value of the test statistic, carry out the test at the 5% significance level.
 one year ago

orderBest ResponseYou've already chosen the best response.0
The test statistic, z, has a value of 1.786 correct to 3 decimal places
 one year ago

glgan1Best ResponseYou've already chosen the best response.1
wow i did this in my A Level which was 1 year ago. damn couldn't remember much. LOL
 one year ago

orderBest ResponseYou've already chosen the best response.0
I'm doing that now... and I never had a teacher for this stuff.... So I'm going to fail :( But I'm doing it again next year, since I'm just trying to push myself to do it in one... :( Do you remember this?
 one year ago

glgan1Best ResponseYou've already chosen the best response.1
Let me look thru my notes.
 one year ago

glgan1Best ResponseYou've already chosen the best response.1
Null hypothesis : population mean=85 Alternative hypothesis: population mean>85(hoping that height will increase) Reject null hypothesis if Z calculated >1.645. \[z=(x\mu)/\sqrt{\sigma^2/n}\] \[x=85.7, \mu=85, \sigma = 4.8, n=150\] you will get Z=1.786(>1.645). Therefore, we reject the nnull hypothesis and the height is shown increased.
 one year ago

orderBest ResponseYou've already chosen the best response.0
How do you get the (>1.645) I understand the others...
 one year ago

glgan1Best ResponseYou've already chosen the best response.1
Because it's a 5% significance level/ 0.05. If we look from the chart about normal distribution, we need to find p equals to 10.05=0.95 to get the Z value. so when p=0.95, z =1.645. And since they want us to prove that whether the height has increased, we will use an upper tail test, therefore '>' sign.
 one year ago

orderBest ResponseYou've already chosen the best response.0
I\ll write more questions soon. If you can, please help! :D I love your explanations!
 one year ago

glgan1Best ResponseYou've already chosen the best response.1
you're welcome. I will try my best. :) My statistics level is just basic, hopefully you can understand what i explained. XD
 one year ago
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