Test for single Mean

Test for single Mean

    1.Form the null hypothesis
    • Ho: µ=µo
    • (i.e) There is no significance difference between the sample mean and the population mean
    2.Form the Alternate hypothesis
    • H1 : µ≠µo (or µ>µo or µ<µo)
    • ie., There is significance difference between the sample mean and the population mean
    3. Level of Significance
    • The level may be fixed at either 5% or 1%
    4. Test statistic
    twhich follows t distribution with (n-1) degrees of freedom
    where And
    6.Find the table value of t corresponding to (n-1) d.f. and the specified level of significance.

    7.Inference
    • If t < ttab we accept the null hypothesis H0. We conclude that there is no significant difference sample mean and population mean (or) if t > ttab we reject the null hypothesis H0. (ie) we accept the alternative hypothesis and conclude that there is significant difference between the sample mean and the population mean
    Example 1
    • Based on field experiments, a new variety of green gram is expected to given a yield of 12.0 quintals per hectare. The variety was tested on 10 randomly selected farmer’s fields. The yield (quintals/hectare) were recorded as 14.3,12.6,13.7,10.9,13.7,12.0,11.4,12.0,12.6,13.1. Do the results conform to the expectation?
    Solution
    • Null hypothesis H0: µ=12.0
    • (i.e) the average yield of the new variety of green gram is 12.0 quintals/hectare.
    • Alternative Hypothesis: H1:µ≠ 12.0
    • (i.e) the average yield is not 12.0 quintals/hectare, it may be less or more than 12 quintals / hectare
    • Level of significance : 5 %
    Test statistic
    t=
    From the given data
    Ex=Ex2
    X
    s=160510.6
    = 1.0853
    s/n
    Now t=
    t=
    • Table value for t corresponding to 5% level of significance and 9 d.f. is 2.262 (two tailed test)

    Inference:

    • t < ttab
    • We accept the null hypothesis H0
    • We conclude that the new variety of green gram will give an average yield of 12 quintals/hectare.
    • Note
    • Before applying t test in case of two samples the equality of their variances has to be tested by using F-test
    Fif
    or
    Fif
    • where s is the variance of the first sample whose size is n1.
    • s is the variance of the second sample whose size is n2.
    • It may be noted that the numerator is always the greater variance. The critical value for F is read from the F table corresponding to a specified d.f. and level of significance

    Inference

    • F <Ftab
    • We accept the null hypothesis H0.(i.e) the variances are equal otherwise the variances are unequal.

Last modified: Monday, 19 March 2012, 9:22 PM