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.

Using this value of the test statistic, carry out the test at the 5% significance level.

The test statistic, z, has a value of 1.786 correct to 3 decimal places

@glgan1

wow i did this in my A Level which was 1 year ago. damn couldn't remember much. LOL

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?

Let me look thru my notes.

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.

How do you get the (>1.645) I understand the others...

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 1-0.05=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.

Ah Ok. Thanks :D

I\ll write more questions soon. If you can, please help! :D I love your explanations!

you're welcome. I will try my best. :) My statistics level is just basic, hopefully you can understand what i explained. XD

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